Bounds on the variance of the log of a positive random variable
0
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Let $X>0$ be a random variable with expectation $E(X)$ and variance $Var(X)$. Can I use this information to get an exact value or upper or lower bounds on the variance of $log X$?
probability probability-distributions random-variables logarithms
asked Dec 18 '18 at 8:19
Alice SchwarzeAlice Schwarze
9114
$endgroup$
add a comment |
0
$begingroup$
Let $X>0$ be a random variable with expectation $E(X)$ and variance $Var(X)$. Can I use this information to get an exact value or upper or lower bounds on the variance of $log X$?
probability probability-distributions random-variables logarithms
asked Dec 18 '18 at 8:19
Alice SchwarzeAlice Schwarze
9114
$endgroup$
$begingroup$
As a start, you might consider a uniform distribution on $(0,2)$ against a random variable taking values $1 pm frac1{sqrt{3}}$ with equal probabilities
$endgroup$
– Henry
Dec 18 '18 at 8:44
$begingroup$
Try using the MGF of $Y = log X$. It is given by $mathbb{E}[e^{tY}] = mathbb{E}[X^t]$.You can substitute $t=1$ and $t=2$ and start to get some bounds. This seems viable for the $X>1$ (i.e. $Y>0$) case, but perhaps less so for $X>0$.
$endgroup$
– Aditya Dua
Dec 18 '18 at 20:36
$begingroup$
@Henry There are several threads on SE concerning the uniform distribution. I looked at them, but I don't see how they can be extended to the case of non-uniform distributions.
$endgroup$
– Alice Schwarze
Dec 18 '18 at 21:18
$begingroup$
@AdityaDua Thank you. In my case, I don't have the MGF of Y or X --- only the first two moments of X. However, out of interest, where do you see problems arising for $0<X<1$?
$endgroup$
– Alice Schwarze
Dec 18 '18 at 21:21
$begingroup$
@AliceSchwarze The MGF is a function of the all the moments. Since you know the first and second moments, you can create upper/lower bounds using the first three terms of the MGF. The other observation to exploit is that the MGF of $Y$, which is by definition $mathbb{E}[e^{tY}]$ can be written as $mathbb{E}[X^t]$. If you set $t=1$ or $t=2$, you will get the first and second moments of $X$. I think you can see where I am going with this.
$endgroup$
– Aditya Dua
Dec 18 '18 at 22:24
add a comment |
0
0
0
$begingroup$
Let $X>0$ be a random variable with expectation $E(X)$ and variance $Var(X)$. Can I use this information to get an exact value or upper or lower bounds on the variance of $log X$?
probability probability-distributions random-variables logarithms
asked Dec 18 '18 at 8:19
Alice SchwarzeAlice Schwarze
9114
$endgroup$
Let $X>0$ be a random variable with expectation $E(X)$ and variance $Var(X)$. Can I use this information to get an exact value or upper or lower bounds on the variance of $log X$?
probability probability-distributions random-variables logarithms
probability probability-distributions random-variables logarithms
asked Dec 18 '18 at 8:19
Alice SchwarzeAlice Schwarze
9114
asked Dec 18 '18 at 8:19
Alice SchwarzeAlice Schwarze
9114
asked Dec 18 '18 at 8:19
Alice SchwarzeAlice Schwarze
9114
asked Dec 18 '18 at 8:19
Alice SchwarzeAlice Schwarze
9114
asked Dec 18 '18 at 8:19
Alice SchwarzeAlice Schwarze
9114
9114
$begingroup$
As a start, you might consider a uniform distribution on $(0,2)$ against a random variable taking values $1 pm frac1{sqrt{3}}$ with equal probabilities
$endgroup$
– Henry
Dec 18 '18 at 8:44
$begingroup$
Try using the MGF of $Y = log X$. It is given by $mathbb{E}[e^{tY}] = mathbb{E}[X^t]$.You can substitute $t=1$ and $t=2$ and start to get some bounds. This seems viable for the $X>1$ (i.e. $Y>0$) case, but perhaps less so for $X>0$.
$endgroup$
– Aditya Dua
Dec 18 '18 at 20:36
$begingroup$
@Henry There are several threads on SE concerning the uniform distribution. I looked at them, but I don't see how they can be extended to the case of non-uniform distributions.
$endgroup$
– Alice Schwarze
Dec 18 '18 at 21:18
$begingroup$
@AdityaDua Thank you. In my case, I don't have the MGF of Y or X --- only the first two moments of X. However, out of interest, where do you see problems arising for $0<X<1$?
$endgroup$
– Alice Schwarze
Dec 18 '18 at 21:21
$begingroup$
@AliceSchwarze The MGF is a function of the all the moments. Since you know the first and second moments, you can create upper/lower bounds using the first three terms of the MGF. The other observation to exploit is that the MGF of $Y$, which is by definition $mathbb{E}[e^{tY}]$ can be written as $mathbb{E}[X^t]$. If you set $t=1$ or $t=2$, you will get the first and second moments of $X$. I think you can see where I am going with this.
$endgroup$
– Aditya Dua
Dec 18 '18 at 22:24
add a comment |
$begingroup$
As a start, you might consider a uniform distribution on $(0,2)$ against a random variable taking values $1 pm frac1{sqrt{3}}$ with equal probabilities
$endgroup$
– Henry
Dec 18 '18 at 8:44
$begingroup$
Try using the MGF of $Y = log X$. It is given by $mathbb{E}[e^{tY}] = mathbb{E}[X^t]$.You can substitute $t=1$ and $t=2$ and start to get some bounds. This seems viable for the $X>1$ (i.e. $Y>0$) case, but perhaps less so for $X>0$.
$endgroup$
– Aditya Dua
Dec 18 '18 at 20:36
$begingroup$
@Henry There are several threads on SE concerning the uniform distribution. I looked at them, but I don't see how they can be extended to the case of non-uniform distributions.
$endgroup$
– Alice Schwarze
Dec 18 '18 at 21:18
$begingroup$
@AdityaDua Thank you. In my case, I don't have the MGF of Y or X --- only the first two moments of X. However, out of interest, where do you see problems arising for $0<X<1$?
$endgroup$
– Alice Schwarze
Dec 18 '18 at 21:21
$begingroup$
@AliceSchwarze The MGF is a function of the all the moments. Since you know the first and second moments, you can create upper/lower bounds using the first three terms of the MGF. The other observation to exploit is that the MGF of $Y$, which is by definition $mathbb{E}[e^{tY}]$ can be written as $mathbb{E}[X^t]$. If you set $t=1$ or $t=2$, you will get the first and second moments of $X$. I think you can see where I am going with this.
$endgroup$
– Aditya Dua
Dec 18 '18 at 22:24
$begingroup$
As a start, you might consider a uniform distribution on $(0,2)$ against a random variable taking values $1 pm frac1{sqrt{3}}$ with equal probabilities
$endgroup$
– Henry
Dec 18 '18 at 8:44
$begingroup$
As a start, you might consider a uniform distribution on $(0,2)$ against a random variable taking values $1 pm frac1{sqrt{3}}$ with equal probabilities
$endgroup$
– Henry
Dec 18 '18 at 8:44
$begingroup$
Try using the MGF of $Y = log X$. It is given by $mathbb{E}[e^{tY}] = mathbb{E}[X^t]$.You can substitute $t=1$ and $t=2$ and start to get some bounds. This seems viable for the $X>1$ (i.e. $Y>0$) case, but perhaps less so for $X>0$.
$endgroup$
– Aditya Dua
Dec 18 '18 at 20:36
$begingroup$
Try using the MGF of $Y = log X$. It is given by $mathbb{E}[e^{tY}] = mathbb{E}[X^t]$.You can substitute $t=1$ and $t=2$ and start to get some bounds. This seems viable for the $X>1$ (i.e. $Y>0$) case, but perhaps less so for $X>0$.
$endgroup$
– Aditya Dua
Dec 18 '18 at 20:36
$begingroup$
@Henry There are several threads on SE concerning the uniform distribution. I looked at them, but I don't see how they can be extended to the case of non-uniform distributions.
$endgroup$
– Alice Schwarze
Dec 18 '18 at 21:18
$begingroup$
@Henry There are several threads on SE concerning the uniform distribution. I looked at them, but I don't see how they can be extended to the case of non-uniform distributions.
$endgroup$
– Alice Schwarze
Dec 18 '18 at 21:18
$begingroup$
@AdityaDua Thank you. In my case, I don't have the MGF of Y or X --- only the first two moments of X. However, out of interest, where do you see problems arising for $0<X<1$?
$endgroup$
– Alice Schwarze
Dec 18 '18 at 21:21
$begingroup$
@AdityaDua Thank you. In my case, I don't have the MGF of Y or X --- only the first two moments of X. However, out of interest, where do you see problems arising for $0<X<1$?
$endgroup$
– Alice Schwarze
Dec 18 '18 at 21:21
$begingroup$
@AliceSchwarze The MGF is a function of the all the moments. Since you know the first and second moments, you can create upper/lower bounds using the first three terms of the MGF. The other observation to exploit is that the MGF of $Y$, which is by definition $mathbb{E}[e^{tY}]$ can be written as $mathbb{E}[X^t]$. If you set $t=1$ or $t=2$, you will get the first and second moments of $X$. I think you can see where I am going with this.
$endgroup$
– Aditya Dua
Dec 18 '18 at 22:24
$begingroup$
@AliceSchwarze The MGF is a function of the all the moments. Since you know the first and second moments, you can create upper/lower bounds using the first three terms of the MGF. The other observation to exploit is that the MGF of $Y$, which is by definition $mathbb{E}[e^{tY}]$ can be written as $mathbb{E}[X^t]$. If you set $t=1$ or $t=2$, you will get the first and second moments of $X$. I think you can see where I am going with this.
$endgroup$
– Aditya Dua
Dec 18 '18 at 22:24
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$begingroup$
As a start, you might consider a uniform distribution on $(0,2)$ against a random variable taking values $1 pm frac1{sqrt{3}}$ with equal probabilities
$endgroup$
– Henry
Dec 18 '18 at 8:44
$begingroup$
Try using the MGF of $Y = log X$. It is given by $mathbb{E}[e^{tY}] = mathbb{E}[X^t]$.You can substitute $t=1$ and $t=2$ and start to get some bounds. This seems viable for the $X>1$ (i.e. $Y>0$) case, but perhaps less so for $X>0$.
$endgroup$
– Aditya Dua
Dec 18 '18 at 20:36
$begingroup$
@Henry There are several threads on SE concerning the uniform distribution. I looked at them, but I don't see how they can be extended to the case of non-uniform distributions.
$endgroup$
– Alice Schwarze
Dec 18 '18 at 21:18
$begingroup$
@AdityaDua Thank you. In my case, I don't have the MGF of Y or X --- only the first two moments of X. However, out of interest, where do you see problems arising for $0<X<1$?
$endgroup$
– Alice Schwarze
Dec 18 '18 at 21:21
$begingroup$
@AliceSchwarze The MGF is a function of the all the moments. Since you know the first and second moments, you can create upper/lower bounds using the first three terms of the MGF. The other observation to exploit is that the MGF of $Y$, which is by definition $mathbb{E}[e^{tY}]$ can be written as $mathbb{E}[X^t]$. If you set $t=1$ or $t=2$, you will get the first and second moments of $X$. I think you can see where I am going with this.
$endgroup$
– Aditya Dua
Dec 18 '18 at 22:24