Inferring observation time from a Brownian motion
$begingroup$
This might be a bit lengthy question. So let me proceed in steps.
General description: I have some observations, based on which I want to infer their occurring time.
Specific setting: Let $W(t)$ be a Brownian motion.
I have an observation $X$ defined as
$$X = BW(delta) + (1-B)W(2delta),$$
where $B$ is a Bernoulli draw with success rate $frac{1}{2}$ and is independent of $W(t)$; and $delta>0$ is some constant observation latency.
In words, the above is meant to capture the setting that I do not know when the observation $X$ occurred:
It might have occurred at time $t=delta$ or at time $t=2delta$, equally likely.
So given the observation $X$, I can infer something about the time of its occurrence: By Bayesian rule,
$$mathbb{P}(B=1|X) = frac{mathbb{P}(B=1)mathbb{P}(X|B=1)}{mathbb{P}(B=1)mathbb{P}(X|B=1)+mathbb{P}(B=0)mathbb{P}(X|B=0)}
= frac{frac{1}{2}mathbb{P}(W(delta)=X)}{frac{1}{2}mathbb{P}(W(delta)=X)+frac{1}{2}mathbb{P}(W(2delta)=X)}.
$$
Given that $W(t)$ is a Brownian motion, $W(delta)$ and $W(2delta)$ are normally distributed with respective densities
$$mathbb{P}(W(delta)=X) = frac{1}{sqrt{2pidelta}}e^{-X^2/(2delta)}$$
and
$$mathbb{P}(W(2delta)=X) = frac{1}{sqrt{4pidelta}}e^{-X^2/(4delta)}.$$
So the above translates to a function of the density ratio
$$mathbb{P}(B=1|X) = frac{1}{1+frac{1}{sqrt{2}}e^{X^2/(4delta)}},
$$
which is decreasing in the magnitude of $|X|$.
That is, the more extreme $X$ realizes to be, the less likely it comes from the early $W(delta)$.
This makes sense as $W(delta)$ is less volatile than $W(2delta)$.
Next I want to consider what happens when $deltadownarrow0$, i.e., when the gap between the timing of the observation asymptotically vanishes.
Intuitively, when $deltadownarrow0$, $W(delta)=W(2delta)=W(0)$ and $X=W(0)$, leaving no room for inference for $B$.
Analytically, however, from the above expression, in the limit of $deltadownarrow0$, $mathbb{P}(B=1|X)=0$, suggesting that I should always infer that $B=0$.
So what is the correct result? Can I infer anything about $B$ in the limit of $deltadownarrow0$?
brownian-motion statistical-inference bayesian
$endgroup$
add a comment |
$begingroup$
This might be a bit lengthy question. So let me proceed in steps.
General description: I have some observations, based on which I want to infer their occurring time.
Specific setting: Let $W(t)$ be a Brownian motion.
I have an observation $X$ defined as
$$X = BW(delta) + (1-B)W(2delta),$$
where $B$ is a Bernoulli draw with success rate $frac{1}{2}$ and is independent of $W(t)$; and $delta>0$ is some constant observation latency.
In words, the above is meant to capture the setting that I do not know when the observation $X$ occurred:
It might have occurred at time $t=delta$ or at time $t=2delta$, equally likely.
So given the observation $X$, I can infer something about the time of its occurrence: By Bayesian rule,
$$mathbb{P}(B=1|X) = frac{mathbb{P}(B=1)mathbb{P}(X|B=1)}{mathbb{P}(B=1)mathbb{P}(X|B=1)+mathbb{P}(B=0)mathbb{P}(X|B=0)}
= frac{frac{1}{2}mathbb{P}(W(delta)=X)}{frac{1}{2}mathbb{P}(W(delta)=X)+frac{1}{2}mathbb{P}(W(2delta)=X)}.
$$
Given that $W(t)$ is a Brownian motion, $W(delta)$ and $W(2delta)$ are normally distributed with respective densities
$$mathbb{P}(W(delta)=X) = frac{1}{sqrt{2pidelta}}e^{-X^2/(2delta)}$$
and
$$mathbb{P}(W(2delta)=X) = frac{1}{sqrt{4pidelta}}e^{-X^2/(4delta)}.$$
So the above translates to a function of the density ratio
$$mathbb{P}(B=1|X) = frac{1}{1+frac{1}{sqrt{2}}e^{X^2/(4delta)}},
$$
which is decreasing in the magnitude of $|X|$.
That is, the more extreme $X$ realizes to be, the less likely it comes from the early $W(delta)$.
This makes sense as $W(delta)$ is less volatile than $W(2delta)$.
Next I want to consider what happens when $deltadownarrow0$, i.e., when the gap between the timing of the observation asymptotically vanishes.
Intuitively, when $deltadownarrow0$, $W(delta)=W(2delta)=W(0)$ and $X=W(0)$, leaving no room for inference for $B$.
Analytically, however, from the above expression, in the limit of $deltadownarrow0$, $mathbb{P}(B=1|X)=0$, suggesting that I should always infer that $B=0$.
So what is the correct result? Can I infer anything about $B$ in the limit of $deltadownarrow0$?
brownian-motion statistical-inference bayesian
$endgroup$
$begingroup$
But the expected size of $X^2$ will also shrink as $delta rightarrow 0$. For $X sim W(delta), mathbb{E}(X^2) = Var(X) = delta$ and for $X sim W(2delta), mathbb{E}(X^2) = Var(X) = 4delta$. So $mathbb{P}(B = 1 mid X)$ will still depend on whether $X^2$ is large compared to $delta$.
$endgroup$
– Alex
Dec 28 '18 at 11:48
$begingroup$
I made a mistake above: for $X sim W(2delta), mathbb{E}(X^2) = 2delta$.
$endgroup$
– Alex
Dec 28 '18 at 11:57
$begingroup$
Right, $mathbb{E}[X^2]=delta$ or $2delta$. That is, on average, the size of $X$ is shrinking too. But for any realization of $X$, the above derivation is correct, I guess?
$endgroup$
– J. Doe
Dec 29 '18 at 13:32
$begingroup$
Thanks for asking the follow-up question. I've tried to give a full answer below.
$endgroup$
– Alex
Dec 29 '18 at 17:38
add a comment |
$begingroup$
This might be a bit lengthy question. So let me proceed in steps.
General description: I have some observations, based on which I want to infer their occurring time.
Specific setting: Let $W(t)$ be a Brownian motion.
I have an observation $X$ defined as
$$X = BW(delta) + (1-B)W(2delta),$$
where $B$ is a Bernoulli draw with success rate $frac{1}{2}$ and is independent of $W(t)$; and $delta>0$ is some constant observation latency.
In words, the above is meant to capture the setting that I do not know when the observation $X$ occurred:
It might have occurred at time $t=delta$ or at time $t=2delta$, equally likely.
So given the observation $X$, I can infer something about the time of its occurrence: By Bayesian rule,
$$mathbb{P}(B=1|X) = frac{mathbb{P}(B=1)mathbb{P}(X|B=1)}{mathbb{P}(B=1)mathbb{P}(X|B=1)+mathbb{P}(B=0)mathbb{P}(X|B=0)}
= frac{frac{1}{2}mathbb{P}(W(delta)=X)}{frac{1}{2}mathbb{P}(W(delta)=X)+frac{1}{2}mathbb{P}(W(2delta)=X)}.
$$
Given that $W(t)$ is a Brownian motion, $W(delta)$ and $W(2delta)$ are normally distributed with respective densities
$$mathbb{P}(W(delta)=X) = frac{1}{sqrt{2pidelta}}e^{-X^2/(2delta)}$$
and
$$mathbb{P}(W(2delta)=X) = frac{1}{sqrt{4pidelta}}e^{-X^2/(4delta)}.$$
So the above translates to a function of the density ratio
$$mathbb{P}(B=1|X) = frac{1}{1+frac{1}{sqrt{2}}e^{X^2/(4delta)}},
$$
which is decreasing in the magnitude of $|X|$.
That is, the more extreme $X$ realizes to be, the less likely it comes from the early $W(delta)$.
This makes sense as $W(delta)$ is less volatile than $W(2delta)$.
Next I want to consider what happens when $deltadownarrow0$, i.e., when the gap between the timing of the observation asymptotically vanishes.
Intuitively, when $deltadownarrow0$, $W(delta)=W(2delta)=W(0)$ and $X=W(0)$, leaving no room for inference for $B$.
Analytically, however, from the above expression, in the limit of $deltadownarrow0$, $mathbb{P}(B=1|X)=0$, suggesting that I should always infer that $B=0$.
So what is the correct result? Can I infer anything about $B$ in the limit of $deltadownarrow0$?
brownian-motion statistical-inference bayesian
$endgroup$
This might be a bit lengthy question. So let me proceed in steps.
General description: I have some observations, based on which I want to infer their occurring time.
Specific setting: Let $W(t)$ be a Brownian motion.
I have an observation $X$ defined as
$$X = BW(delta) + (1-B)W(2delta),$$
where $B$ is a Bernoulli draw with success rate $frac{1}{2}$ and is independent of $W(t)$; and $delta>0$ is some constant observation latency.
In words, the above is meant to capture the setting that I do not know when the observation $X$ occurred:
It might have occurred at time $t=delta$ or at time $t=2delta$, equally likely.
So given the observation $X$, I can infer something about the time of its occurrence: By Bayesian rule,
$$mathbb{P}(B=1|X) = frac{mathbb{P}(B=1)mathbb{P}(X|B=1)}{mathbb{P}(B=1)mathbb{P}(X|B=1)+mathbb{P}(B=0)mathbb{P}(X|B=0)}
= frac{frac{1}{2}mathbb{P}(W(delta)=X)}{frac{1}{2}mathbb{P}(W(delta)=X)+frac{1}{2}mathbb{P}(W(2delta)=X)}.
$$
Given that $W(t)$ is a Brownian motion, $W(delta)$ and $W(2delta)$ are normally distributed with respective densities
$$mathbb{P}(W(delta)=X) = frac{1}{sqrt{2pidelta}}e^{-X^2/(2delta)}$$
and
$$mathbb{P}(W(2delta)=X) = frac{1}{sqrt{4pidelta}}e^{-X^2/(4delta)}.$$
So the above translates to a function of the density ratio
$$mathbb{P}(B=1|X) = frac{1}{1+frac{1}{sqrt{2}}e^{X^2/(4delta)}},
$$
which is decreasing in the magnitude of $|X|$.
That is, the more extreme $X$ realizes to be, the less likely it comes from the early $W(delta)$.
This makes sense as $W(delta)$ is less volatile than $W(2delta)$.
Next I want to consider what happens when $deltadownarrow0$, i.e., when the gap between the timing of the observation asymptotically vanishes.
Intuitively, when $deltadownarrow0$, $W(delta)=W(2delta)=W(0)$ and $X=W(0)$, leaving no room for inference for $B$.
Analytically, however, from the above expression, in the limit of $deltadownarrow0$, $mathbb{P}(B=1|X)=0$, suggesting that I should always infer that $B=0$.
So what is the correct result? Can I infer anything about $B$ in the limit of $deltadownarrow0$?
brownian-motion statistical-inference bayesian
brownian-motion statistical-inference bayesian
asked Dec 28 '18 at 6:26
J. DoeJ. Doe
61
61
$begingroup$
But the expected size of $X^2$ will also shrink as $delta rightarrow 0$. For $X sim W(delta), mathbb{E}(X^2) = Var(X) = delta$ and for $X sim W(2delta), mathbb{E}(X^2) = Var(X) = 4delta$. So $mathbb{P}(B = 1 mid X)$ will still depend on whether $X^2$ is large compared to $delta$.
$endgroup$
– Alex
Dec 28 '18 at 11:48
$begingroup$
I made a mistake above: for $X sim W(2delta), mathbb{E}(X^2) = 2delta$.
$endgroup$
– Alex
Dec 28 '18 at 11:57
$begingroup$
Right, $mathbb{E}[X^2]=delta$ or $2delta$. That is, on average, the size of $X$ is shrinking too. But for any realization of $X$, the above derivation is correct, I guess?
$endgroup$
– J. Doe
Dec 29 '18 at 13:32
$begingroup$
Thanks for asking the follow-up question. I've tried to give a full answer below.
$endgroup$
– Alex
Dec 29 '18 at 17:38
add a comment |
$begingroup$
But the expected size of $X^2$ will also shrink as $delta rightarrow 0$. For $X sim W(delta), mathbb{E}(X^2) = Var(X) = delta$ and for $X sim W(2delta), mathbb{E}(X^2) = Var(X) = 4delta$. So $mathbb{P}(B = 1 mid X)$ will still depend on whether $X^2$ is large compared to $delta$.
$endgroup$
– Alex
Dec 28 '18 at 11:48
$begingroup$
I made a mistake above: for $X sim W(2delta), mathbb{E}(X^2) = 2delta$.
$endgroup$
– Alex
Dec 28 '18 at 11:57
$begingroup$
Right, $mathbb{E}[X^2]=delta$ or $2delta$. That is, on average, the size of $X$ is shrinking too. But for any realization of $X$, the above derivation is correct, I guess?
$endgroup$
– J. Doe
Dec 29 '18 at 13:32
$begingroup$
Thanks for asking the follow-up question. I've tried to give a full answer below.
$endgroup$
– Alex
Dec 29 '18 at 17:38
$begingroup$
But the expected size of $X^2$ will also shrink as $delta rightarrow 0$. For $X sim W(delta), mathbb{E}(X^2) = Var(X) = delta$ and for $X sim W(2delta), mathbb{E}(X^2) = Var(X) = 4delta$. So $mathbb{P}(B = 1 mid X)$ will still depend on whether $X^2$ is large compared to $delta$.
$endgroup$
– Alex
Dec 28 '18 at 11:48
$begingroup$
But the expected size of $X^2$ will also shrink as $delta rightarrow 0$. For $X sim W(delta), mathbb{E}(X^2) = Var(X) = delta$ and for $X sim W(2delta), mathbb{E}(X^2) = Var(X) = 4delta$. So $mathbb{P}(B = 1 mid X)$ will still depend on whether $X^2$ is large compared to $delta$.
$endgroup$
– Alex
Dec 28 '18 at 11:48
$begingroup$
I made a mistake above: for $X sim W(2delta), mathbb{E}(X^2) = 2delta$.
$endgroup$
– Alex
Dec 28 '18 at 11:57
$begingroup$
I made a mistake above: for $X sim W(2delta), mathbb{E}(X^2) = 2delta$.
$endgroup$
– Alex
Dec 28 '18 at 11:57
$begingroup$
Right, $mathbb{E}[X^2]=delta$ or $2delta$. That is, on average, the size of $X$ is shrinking too. But for any realization of $X$, the above derivation is correct, I guess?
$endgroup$
– J. Doe
Dec 29 '18 at 13:32
$begingroup$
Right, $mathbb{E}[X^2]=delta$ or $2delta$. That is, on average, the size of $X$ is shrinking too. But for any realization of $X$, the above derivation is correct, I guess?
$endgroup$
– J. Doe
Dec 29 '18 at 13:32
$begingroup$
Thanks for asking the follow-up question. I've tried to give a full answer below.
$endgroup$
– Alex
Dec 29 '18 at 17:38
$begingroup$
Thanks for asking the follow-up question. I've tried to give a full answer below.
$endgroup$
– Alex
Dec 29 '18 at 17:38
add a comment |
1 Answer
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oldest
votes
$begingroup$
The probability $$mathbb{P}(B = 1 mid X) = frac{1}{1 + frac{1}{sqrt{2}}e^{X^2/(4delta)}}$$ only depends on $X$ and $delta$ through the term $X^2/(4delta)$. Now let's think about the cases when $B = 1$ or $B = 0$.
When $B = 1$ then $X sim N(0, delta)$, so $X/sqrt{delta} sim N(0, 1)$. Therefore $X^2/delta sim chi^2_1$, a chi-squared distribution. And $X^2/(4delta) sim chi^2_1/4$, the distribution of a $chi^2_1$ random variable divided by $4$.
When $B = 0$ then $X sim N(0, 2delta)$, so $X/sqrt{2delta} sim N(0, 1)$. Therefore $X^2/(2delta) sim chi^2_1$, and $X^2/(4delta) sim chi^2_1/2$, the distribution of a $chi^2_1$ random variable divided by $2$.
In either case, the distribution of $X^2/(4delta)$ does not depend on $delta$. So as $delta rightarrow 0$, $mathbb{P}(B = 1mid X)$ does not tend to any particular limit.
$endgroup$
$begingroup$
Thanks a lot, Alex. This is very clear for the conclusion that "the distribution of $X^2/(4delta)$ does not depend on $delta$". But I guess in evaluating $mathbb{P}(B=1|X)$, it is the realization of $X$ that matters (as opposed to its distribution)?
$endgroup$
– J. Doe
Dec 30 '18 at 10:50
$begingroup$
I agree, what I'm saying is that $mathbb{P}(B = 1 mid X)$ is not dependent on $delta$. For any value of $delta$ the probability that $B = 1$ given $X$ just depends on whether the realization of $X^2/(4delta)$ is larger or smaller than we would expect if $B = 1$. As $delta downarrow 0$, nothing of interest happens to the probability we will assign to the event $B = 1$.
$endgroup$
– Alex
Dec 30 '18 at 13:24
$begingroup$
Thanks again, Alex. Very much appreciated. So I think I see the issue much clearer now but I am still not so sure of the conclusion you hinted above. Indeed, fixing any value of $delta$, $mathbb{P}(B=1|X)$ just depends on the realization of $X^2/delta$. So as I take the limit of $deltadownarrow0$, the realization of $X^2/delta$ explodes to infinity (for any $|X|neq0$) and this affects $mathbb{P}(B=1|X)$, no?
$endgroup$
– J. Doe
Dec 31 '18 at 0:19
$begingroup$
Sure, if you fix $X neq 0$ then $X^2/delta rightarrow infty$ and $mathbb{P}(B = 1 mid X) rightarrow 0$ as $delta downarrow 0$. I assumed you were talking about calculating $mathbb{P}(B = 1 mid X)$ when $X=BW(delta)+(1−B)W(2delta)$. So if you were thinking about this probability as $delta downarrow 0$, you would need a sequence of $delta$s $delta_0, delta_1, dots$ that converges down to $0$, and for each $delta_i$ you need a new random variable $X_i =BW(delta_i)+(1−B)W(2delta_i)$. Then the sequence of $mathbb{P}(B = 1 mid X_i)$ realizations will not tend to a limit.
$endgroup$
– Alex
Dec 31 '18 at 8:30
$begingroup$
Compare with the Weak Law of Large Numbers, where a probability depends on a constant $n$ and converges to $1$ as $n rightarrow infty$ en.wikipedia.org/wiki/Law_of_large_numbers#Weak_law
$endgroup$
– Alex
Dec 31 '18 at 9:57
add a comment |
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$begingroup$
The probability $$mathbb{P}(B = 1 mid X) = frac{1}{1 + frac{1}{sqrt{2}}e^{X^2/(4delta)}}$$ only depends on $X$ and $delta$ through the term $X^2/(4delta)$. Now let's think about the cases when $B = 1$ or $B = 0$.
When $B = 1$ then $X sim N(0, delta)$, so $X/sqrt{delta} sim N(0, 1)$. Therefore $X^2/delta sim chi^2_1$, a chi-squared distribution. And $X^2/(4delta) sim chi^2_1/4$, the distribution of a $chi^2_1$ random variable divided by $4$.
When $B = 0$ then $X sim N(0, 2delta)$, so $X/sqrt{2delta} sim N(0, 1)$. Therefore $X^2/(2delta) sim chi^2_1$, and $X^2/(4delta) sim chi^2_1/2$, the distribution of a $chi^2_1$ random variable divided by $2$.
In either case, the distribution of $X^2/(4delta)$ does not depend on $delta$. So as $delta rightarrow 0$, $mathbb{P}(B = 1mid X)$ does not tend to any particular limit.
$endgroup$
$begingroup$
Thanks a lot, Alex. This is very clear for the conclusion that "the distribution of $X^2/(4delta)$ does not depend on $delta$". But I guess in evaluating $mathbb{P}(B=1|X)$, it is the realization of $X$ that matters (as opposed to its distribution)?
$endgroup$
– J. Doe
Dec 30 '18 at 10:50
$begingroup$
I agree, what I'm saying is that $mathbb{P}(B = 1 mid X)$ is not dependent on $delta$. For any value of $delta$ the probability that $B = 1$ given $X$ just depends on whether the realization of $X^2/(4delta)$ is larger or smaller than we would expect if $B = 1$. As $delta downarrow 0$, nothing of interest happens to the probability we will assign to the event $B = 1$.
$endgroup$
– Alex
Dec 30 '18 at 13:24
$begingroup$
Thanks again, Alex. Very much appreciated. So I think I see the issue much clearer now but I am still not so sure of the conclusion you hinted above. Indeed, fixing any value of $delta$, $mathbb{P}(B=1|X)$ just depends on the realization of $X^2/delta$. So as I take the limit of $deltadownarrow0$, the realization of $X^2/delta$ explodes to infinity (for any $|X|neq0$) and this affects $mathbb{P}(B=1|X)$, no?
$endgroup$
– J. Doe
Dec 31 '18 at 0:19
$begingroup$
Sure, if you fix $X neq 0$ then $X^2/delta rightarrow infty$ and $mathbb{P}(B = 1 mid X) rightarrow 0$ as $delta downarrow 0$. I assumed you were talking about calculating $mathbb{P}(B = 1 mid X)$ when $X=BW(delta)+(1−B)W(2delta)$. So if you were thinking about this probability as $delta downarrow 0$, you would need a sequence of $delta$s $delta_0, delta_1, dots$ that converges down to $0$, and for each $delta_i$ you need a new random variable $X_i =BW(delta_i)+(1−B)W(2delta_i)$. Then the sequence of $mathbb{P}(B = 1 mid X_i)$ realizations will not tend to a limit.
$endgroup$
– Alex
Dec 31 '18 at 8:30
$begingroup$
Compare with the Weak Law of Large Numbers, where a probability depends on a constant $n$ and converges to $1$ as $n rightarrow infty$ en.wikipedia.org/wiki/Law_of_large_numbers#Weak_law
$endgroup$
– Alex
Dec 31 '18 at 9:57
add a comment |
$begingroup$
The probability $$mathbb{P}(B = 1 mid X) = frac{1}{1 + frac{1}{sqrt{2}}e^{X^2/(4delta)}}$$ only depends on $X$ and $delta$ through the term $X^2/(4delta)$. Now let's think about the cases when $B = 1$ or $B = 0$.
When $B = 1$ then $X sim N(0, delta)$, so $X/sqrt{delta} sim N(0, 1)$. Therefore $X^2/delta sim chi^2_1$, a chi-squared distribution. And $X^2/(4delta) sim chi^2_1/4$, the distribution of a $chi^2_1$ random variable divided by $4$.
When $B = 0$ then $X sim N(0, 2delta)$, so $X/sqrt{2delta} sim N(0, 1)$. Therefore $X^2/(2delta) sim chi^2_1$, and $X^2/(4delta) sim chi^2_1/2$, the distribution of a $chi^2_1$ random variable divided by $2$.
In either case, the distribution of $X^2/(4delta)$ does not depend on $delta$. So as $delta rightarrow 0$, $mathbb{P}(B = 1mid X)$ does not tend to any particular limit.
$endgroup$
$begingroup$
Thanks a lot, Alex. This is very clear for the conclusion that "the distribution of $X^2/(4delta)$ does not depend on $delta$". But I guess in evaluating $mathbb{P}(B=1|X)$, it is the realization of $X$ that matters (as opposed to its distribution)?
$endgroup$
– J. Doe
Dec 30 '18 at 10:50
$begingroup$
I agree, what I'm saying is that $mathbb{P}(B = 1 mid X)$ is not dependent on $delta$. For any value of $delta$ the probability that $B = 1$ given $X$ just depends on whether the realization of $X^2/(4delta)$ is larger or smaller than we would expect if $B = 1$. As $delta downarrow 0$, nothing of interest happens to the probability we will assign to the event $B = 1$.
$endgroup$
– Alex
Dec 30 '18 at 13:24
$begingroup$
Thanks again, Alex. Very much appreciated. So I think I see the issue much clearer now but I am still not so sure of the conclusion you hinted above. Indeed, fixing any value of $delta$, $mathbb{P}(B=1|X)$ just depends on the realization of $X^2/delta$. So as I take the limit of $deltadownarrow0$, the realization of $X^2/delta$ explodes to infinity (for any $|X|neq0$) and this affects $mathbb{P}(B=1|X)$, no?
$endgroup$
– J. Doe
Dec 31 '18 at 0:19
$begingroup$
Sure, if you fix $X neq 0$ then $X^2/delta rightarrow infty$ and $mathbb{P}(B = 1 mid X) rightarrow 0$ as $delta downarrow 0$. I assumed you were talking about calculating $mathbb{P}(B = 1 mid X)$ when $X=BW(delta)+(1−B)W(2delta)$. So if you were thinking about this probability as $delta downarrow 0$, you would need a sequence of $delta$s $delta_0, delta_1, dots$ that converges down to $0$, and for each $delta_i$ you need a new random variable $X_i =BW(delta_i)+(1−B)W(2delta_i)$. Then the sequence of $mathbb{P}(B = 1 mid X_i)$ realizations will not tend to a limit.
$endgroup$
– Alex
Dec 31 '18 at 8:30
$begingroup$
Compare with the Weak Law of Large Numbers, where a probability depends on a constant $n$ and converges to $1$ as $n rightarrow infty$ en.wikipedia.org/wiki/Law_of_large_numbers#Weak_law
$endgroup$
– Alex
Dec 31 '18 at 9:57
add a comment |
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The probability $$mathbb{P}(B = 1 mid X) = frac{1}{1 + frac{1}{sqrt{2}}e^{X^2/(4delta)}}$$ only depends on $X$ and $delta$ through the term $X^2/(4delta)$. Now let's think about the cases when $B = 1$ or $B = 0$.
When $B = 1$ then $X sim N(0, delta)$, so $X/sqrt{delta} sim N(0, 1)$. Therefore $X^2/delta sim chi^2_1$, a chi-squared distribution. And $X^2/(4delta) sim chi^2_1/4$, the distribution of a $chi^2_1$ random variable divided by $4$.
When $B = 0$ then $X sim N(0, 2delta)$, so $X/sqrt{2delta} sim N(0, 1)$. Therefore $X^2/(2delta) sim chi^2_1$, and $X^2/(4delta) sim chi^2_1/2$, the distribution of a $chi^2_1$ random variable divided by $2$.
In either case, the distribution of $X^2/(4delta)$ does not depend on $delta$. So as $delta rightarrow 0$, $mathbb{P}(B = 1mid X)$ does not tend to any particular limit.
$endgroup$
The probability $$mathbb{P}(B = 1 mid X) = frac{1}{1 + frac{1}{sqrt{2}}e^{X^2/(4delta)}}$$ only depends on $X$ and $delta$ through the term $X^2/(4delta)$. Now let's think about the cases when $B = 1$ or $B = 0$.
When $B = 1$ then $X sim N(0, delta)$, so $X/sqrt{delta} sim N(0, 1)$. Therefore $X^2/delta sim chi^2_1$, a chi-squared distribution. And $X^2/(4delta) sim chi^2_1/4$, the distribution of a $chi^2_1$ random variable divided by $4$.
When $B = 0$ then $X sim N(0, 2delta)$, so $X/sqrt{2delta} sim N(0, 1)$. Therefore $X^2/(2delta) sim chi^2_1$, and $X^2/(4delta) sim chi^2_1/2$, the distribution of a $chi^2_1$ random variable divided by $2$.
In either case, the distribution of $X^2/(4delta)$ does not depend on $delta$. So as $delta rightarrow 0$, $mathbb{P}(B = 1mid X)$ does not tend to any particular limit.
answered Dec 29 '18 at 17:38
AlexAlex
734412
734412
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Thanks a lot, Alex. This is very clear for the conclusion that "the distribution of $X^2/(4delta)$ does not depend on $delta$". But I guess in evaluating $mathbb{P}(B=1|X)$, it is the realization of $X$ that matters (as opposed to its distribution)?
$endgroup$
– J. Doe
Dec 30 '18 at 10:50
$begingroup$
I agree, what I'm saying is that $mathbb{P}(B = 1 mid X)$ is not dependent on $delta$. For any value of $delta$ the probability that $B = 1$ given $X$ just depends on whether the realization of $X^2/(4delta)$ is larger or smaller than we would expect if $B = 1$. As $delta downarrow 0$, nothing of interest happens to the probability we will assign to the event $B = 1$.
$endgroup$
– Alex
Dec 30 '18 at 13:24
$begingroup$
Thanks again, Alex. Very much appreciated. So I think I see the issue much clearer now but I am still not so sure of the conclusion you hinted above. Indeed, fixing any value of $delta$, $mathbb{P}(B=1|X)$ just depends on the realization of $X^2/delta$. So as I take the limit of $deltadownarrow0$, the realization of $X^2/delta$ explodes to infinity (for any $|X|neq0$) and this affects $mathbb{P}(B=1|X)$, no?
$endgroup$
– J. Doe
Dec 31 '18 at 0:19
$begingroup$
Sure, if you fix $X neq 0$ then $X^2/delta rightarrow infty$ and $mathbb{P}(B = 1 mid X) rightarrow 0$ as $delta downarrow 0$. I assumed you were talking about calculating $mathbb{P}(B = 1 mid X)$ when $X=BW(delta)+(1−B)W(2delta)$. So if you were thinking about this probability as $delta downarrow 0$, you would need a sequence of $delta$s $delta_0, delta_1, dots$ that converges down to $0$, and for each $delta_i$ you need a new random variable $X_i =BW(delta_i)+(1−B)W(2delta_i)$. Then the sequence of $mathbb{P}(B = 1 mid X_i)$ realizations will not tend to a limit.
$endgroup$
– Alex
Dec 31 '18 at 8:30
$begingroup$
Compare with the Weak Law of Large Numbers, where a probability depends on a constant $n$ and converges to $1$ as $n rightarrow infty$ en.wikipedia.org/wiki/Law_of_large_numbers#Weak_law
$endgroup$
– Alex
Dec 31 '18 at 9:57
add a comment |
$begingroup$
Thanks a lot, Alex. This is very clear for the conclusion that "the distribution of $X^2/(4delta)$ does not depend on $delta$". But I guess in evaluating $mathbb{P}(B=1|X)$, it is the realization of $X$ that matters (as opposed to its distribution)?
$endgroup$
– J. Doe
Dec 30 '18 at 10:50
$begingroup$
I agree, what I'm saying is that $mathbb{P}(B = 1 mid X)$ is not dependent on $delta$. For any value of $delta$ the probability that $B = 1$ given $X$ just depends on whether the realization of $X^2/(4delta)$ is larger or smaller than we would expect if $B = 1$. As $delta downarrow 0$, nothing of interest happens to the probability we will assign to the event $B = 1$.
$endgroup$
– Alex
Dec 30 '18 at 13:24
$begingroup$
Thanks again, Alex. Very much appreciated. So I think I see the issue much clearer now but I am still not so sure of the conclusion you hinted above. Indeed, fixing any value of $delta$, $mathbb{P}(B=1|X)$ just depends on the realization of $X^2/delta$. So as I take the limit of $deltadownarrow0$, the realization of $X^2/delta$ explodes to infinity (for any $|X|neq0$) and this affects $mathbb{P}(B=1|X)$, no?
$endgroup$
– J. Doe
Dec 31 '18 at 0:19
$begingroup$
Sure, if you fix $X neq 0$ then $X^2/delta rightarrow infty$ and $mathbb{P}(B = 1 mid X) rightarrow 0$ as $delta downarrow 0$. I assumed you were talking about calculating $mathbb{P}(B = 1 mid X)$ when $X=BW(delta)+(1−B)W(2delta)$. So if you were thinking about this probability as $delta downarrow 0$, you would need a sequence of $delta$s $delta_0, delta_1, dots$ that converges down to $0$, and for each $delta_i$ you need a new random variable $X_i =BW(delta_i)+(1−B)W(2delta_i)$. Then the sequence of $mathbb{P}(B = 1 mid X_i)$ realizations will not tend to a limit.
$endgroup$
– Alex
Dec 31 '18 at 8:30
$begingroup$
Compare with the Weak Law of Large Numbers, where a probability depends on a constant $n$ and converges to $1$ as $n rightarrow infty$ en.wikipedia.org/wiki/Law_of_large_numbers#Weak_law
$endgroup$
– Alex
Dec 31 '18 at 9:57
$begingroup$
Thanks a lot, Alex. This is very clear for the conclusion that "the distribution of $X^2/(4delta)$ does not depend on $delta$". But I guess in evaluating $mathbb{P}(B=1|X)$, it is the realization of $X$ that matters (as opposed to its distribution)?
$endgroup$
– J. Doe
Dec 30 '18 at 10:50
$begingroup$
Thanks a lot, Alex. This is very clear for the conclusion that "the distribution of $X^2/(4delta)$ does not depend on $delta$". But I guess in evaluating $mathbb{P}(B=1|X)$, it is the realization of $X$ that matters (as opposed to its distribution)?
$endgroup$
– J. Doe
Dec 30 '18 at 10:50
$begingroup$
I agree, what I'm saying is that $mathbb{P}(B = 1 mid X)$ is not dependent on $delta$. For any value of $delta$ the probability that $B = 1$ given $X$ just depends on whether the realization of $X^2/(4delta)$ is larger or smaller than we would expect if $B = 1$. As $delta downarrow 0$, nothing of interest happens to the probability we will assign to the event $B = 1$.
$endgroup$
– Alex
Dec 30 '18 at 13:24
$begingroup$
I agree, what I'm saying is that $mathbb{P}(B = 1 mid X)$ is not dependent on $delta$. For any value of $delta$ the probability that $B = 1$ given $X$ just depends on whether the realization of $X^2/(4delta)$ is larger or smaller than we would expect if $B = 1$. As $delta downarrow 0$, nothing of interest happens to the probability we will assign to the event $B = 1$.
$endgroup$
– Alex
Dec 30 '18 at 13:24
$begingroup$
Thanks again, Alex. Very much appreciated. So I think I see the issue much clearer now but I am still not so sure of the conclusion you hinted above. Indeed, fixing any value of $delta$, $mathbb{P}(B=1|X)$ just depends on the realization of $X^2/delta$. So as I take the limit of $deltadownarrow0$, the realization of $X^2/delta$ explodes to infinity (for any $|X|neq0$) and this affects $mathbb{P}(B=1|X)$, no?
$endgroup$
– J. Doe
Dec 31 '18 at 0:19
$begingroup$
Thanks again, Alex. Very much appreciated. So I think I see the issue much clearer now but I am still not so sure of the conclusion you hinted above. Indeed, fixing any value of $delta$, $mathbb{P}(B=1|X)$ just depends on the realization of $X^2/delta$. So as I take the limit of $deltadownarrow0$, the realization of $X^2/delta$ explodes to infinity (for any $|X|neq0$) and this affects $mathbb{P}(B=1|X)$, no?
$endgroup$
– J. Doe
Dec 31 '18 at 0:19
$begingroup$
Sure, if you fix $X neq 0$ then $X^2/delta rightarrow infty$ and $mathbb{P}(B = 1 mid X) rightarrow 0$ as $delta downarrow 0$. I assumed you were talking about calculating $mathbb{P}(B = 1 mid X)$ when $X=BW(delta)+(1−B)W(2delta)$. So if you were thinking about this probability as $delta downarrow 0$, you would need a sequence of $delta$s $delta_0, delta_1, dots$ that converges down to $0$, and for each $delta_i$ you need a new random variable $X_i =BW(delta_i)+(1−B)W(2delta_i)$. Then the sequence of $mathbb{P}(B = 1 mid X_i)$ realizations will not tend to a limit.
$endgroup$
– Alex
Dec 31 '18 at 8:30
$begingroup$
Sure, if you fix $X neq 0$ then $X^2/delta rightarrow infty$ and $mathbb{P}(B = 1 mid X) rightarrow 0$ as $delta downarrow 0$. I assumed you were talking about calculating $mathbb{P}(B = 1 mid X)$ when $X=BW(delta)+(1−B)W(2delta)$. So if you were thinking about this probability as $delta downarrow 0$, you would need a sequence of $delta$s $delta_0, delta_1, dots$ that converges down to $0$, and for each $delta_i$ you need a new random variable $X_i =BW(delta_i)+(1−B)W(2delta_i)$. Then the sequence of $mathbb{P}(B = 1 mid X_i)$ realizations will not tend to a limit.
$endgroup$
– Alex
Dec 31 '18 at 8:30
$begingroup$
Compare with the Weak Law of Large Numbers, where a probability depends on a constant $n$ and converges to $1$ as $n rightarrow infty$ en.wikipedia.org/wiki/Law_of_large_numbers#Weak_law
$endgroup$
– Alex
Dec 31 '18 at 9:57
$begingroup$
Compare with the Weak Law of Large Numbers, where a probability depends on a constant $n$ and converges to $1$ as $n rightarrow infty$ en.wikipedia.org/wiki/Law_of_large_numbers#Weak_law
$endgroup$
– Alex
Dec 31 '18 at 9:57
add a comment |
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$begingroup$
But the expected size of $X^2$ will also shrink as $delta rightarrow 0$. For $X sim W(delta), mathbb{E}(X^2) = Var(X) = delta$ and for $X sim W(2delta), mathbb{E}(X^2) = Var(X) = 4delta$. So $mathbb{P}(B = 1 mid X)$ will still depend on whether $X^2$ is large compared to $delta$.
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– Alex
Dec 28 '18 at 11:48
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I made a mistake above: for $X sim W(2delta), mathbb{E}(X^2) = 2delta$.
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– Alex
Dec 28 '18 at 11:57
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Right, $mathbb{E}[X^2]=delta$ or $2delta$. That is, on average, the size of $X$ is shrinking too. But for any realization of $X$, the above derivation is correct, I guess?
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– J. Doe
Dec 29 '18 at 13:32
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Thanks for asking the follow-up question. I've tried to give a full answer below.
$endgroup$
– Alex
Dec 29 '18 at 17:38