Find Chebyshev's upper bound for this probability
$begingroup$
We are given that $X$ is $Uniform[4,10]$ with mean $7$ and variance $3$.
We are first asked to figure out $P(X leq 5 text { or } X geq 9$)
Well this is a uniform graph with height = $1/(10-4)=1/6$, and we can do the following:
$P(X leq 5 text { or } X geq 9) = P(X leq 5) + P(Xgeq 9)=(5-4)/6+(10-9)/6=2/6=1/3$
Now I am asking the following:
Let $X_1, X_2, X_3, dots, X_{16}$ be a random sample from this uniform distribution. We are asked to find Chebyshev's upper bound for $P(overline{X} leq 5 text { or } overline{X} geq 9)$ where $overline{X}=dfrac{sum_{i=1}^{16}X_i}{16}$
Again, we can break up the probability to be as follows:
$P(overline{X} leq 5) + P(overline{X} geq 9)leq text{upper bound}$
Chebyshev Inequality is the following: $P(|X-mu_x| geq a)leq dfrac{var(x)}{a^2}$
So, for $P(overline{X}leq 5)$, we have $P(|overline{X}-7|leq5)=1-dfrac{3}{25}$
For $P(overline{X}geq 9)$, we have $P(|overline{X}-7|geq 9)=dfrac{3}{9^2}$
So in total, the upper bound must be $1-dfrac{3}{25}+dfrac{3}{9^2}=0.917$
Is this correct? I don't see where we used $overline{X}$ in this. There would be nothing different if I considered $X$, instead of $overline{X}$, so I think I am missing something.
Maybe the $var(x)$ should actually be $sigma^2/n=3/16$. But in my notes, I have that this is only the case if $X_1, X_2, dots, X_n$ are $N(mu, sigma^2)$. Here we have that they are uniform, so I am not sure.
probability statistics inequality
asked Dec 5 '18 at 21:17
K Split XK Split X
4,19611131
$endgroup$
add a comment |
$begingroup$
We are given that $X$ is $Uniform[4,10]$ with mean $7$ and variance $3$.
We are first asked to figure out $P(X leq 5 text { or } X geq 9$)
Well this is a uniform graph with height = $1/(10-4)=1/6$, and we can do the following:
$P(X leq 5 text { or } X geq 9) = P(X leq 5) + P(Xgeq 9)=(5-4)/6+(10-9)/6=2/6=1/3$
Now I am asking the following:
Let $X_1, X_2, X_3, dots, X_{16}$ be a random sample from this uniform distribution. We are asked to find Chebyshev's upper bound for $P(overline{X} leq 5 text { or } overline{X} geq 9)$ where $overline{X}=dfrac{sum_{i=1}^{16}X_i}{16}$
Again, we can break up the probability to be as follows:
$P(overline{X} leq 5) + P(overline{X} geq 9)leq text{upper bound}$
Chebyshev Inequality is the following: $P(|X-mu_x| geq a)leq dfrac{var(x)}{a^2}$
So, for $P(overline{X}leq 5)$, we have $P(|overline{X}-7|leq5)=1-dfrac{3}{25}$
For $P(overline{X}geq 9)$, we have $P(|overline{X}-7|geq 9)=dfrac{3}{9^2}$
So in total, the upper bound must be $1-dfrac{3}{25}+dfrac{3}{9^2}=0.917$
Is this correct? I don't see where we used $overline{X}$ in this. There would be nothing different if I considered $X$, instead of $overline{X}$, so I think I am missing something.
Maybe the $var(x)$ should actually be $sigma^2/n=3/16$. But in my notes, I have that this is only the case if $X_1, X_2, dots, X_n$ are $N(mu, sigma^2)$. Here we have that they are uniform, so I am not sure.
probability statistics inequality
asked Dec 5 '18 at 21:17
K Split XK Split X
4,19611131
$endgroup$
$begingroup$
In general, $text{var}(bar{X}) = frac{1}{n^2} text{var}(sum_{i=1}^n X_i) = frac{1}{n} text{var}(X_1)$ for i.i.d. $X_i$. You should use this variance when applying Chebychev to $bar{X}$.
$endgroup$
– angryavian
Dec 5 '18 at 21:27
add a comment |
$begingroup$
We are given that $X$ is $Uniform[4,10]$ with mean $7$ and variance $3$.
We are first asked to figure out $P(X leq 5 text { or } X geq 9$)
Well this is a uniform graph with height = $1/(10-4)=1/6$, and we can do the following:
$P(X leq 5 text { or } X geq 9) = P(X leq 5) + P(Xgeq 9)=(5-4)/6+(10-9)/6=2/6=1/3$
Now I am asking the following:
Let $X_1, X_2, X_3, dots, X_{16}$ be a random sample from this uniform distribution. We are asked to find Chebyshev's upper bound for $P(overline{X} leq 5 text { or } overline{X} geq 9)$ where $overline{X}=dfrac{sum_{i=1}^{16}X_i}{16}$
Again, we can break up the probability to be as follows:
$P(overline{X} leq 5) + P(overline{X} geq 9)leq text{upper bound}$
Chebyshev Inequality is the following: $P(|X-mu_x| geq a)leq dfrac{var(x)}{a^2}$
So, for $P(overline{X}leq 5)$, we have $P(|overline{X}-7|leq5)=1-dfrac{3}{25}$
For $P(overline{X}geq 9)$, we have $P(|overline{X}-7|geq 9)=dfrac{3}{9^2}$
So in total, the upper bound must be $1-dfrac{3}{25}+dfrac{3}{9^2}=0.917$
Is this correct? I don't see where we used $overline{X}$ in this. There would be nothing different if I considered $X$, instead of $overline{X}$, so I think I am missing something.
Maybe the $var(x)$ should actually be $sigma^2/n=3/16$. But in my notes, I have that this is only the case if $X_1, X_2, dots, X_n$ are $N(mu, sigma^2)$. Here we have that they are uniform, so I am not sure.
probability statistics inequality
asked Dec 5 '18 at 21:17
K Split XK Split X
4,19611131
$endgroup$
We are given that $X$ is $Uniform[4,10]$ with mean $7$ and variance $3$.
We are first asked to figure out $P(X leq 5 text { or } X geq 9$)
Well this is a uniform graph with height = $1/(10-4)=1/6$, and we can do the following:
$P(X leq 5 text { or } X geq 9) = P(X leq 5) + P(Xgeq 9)=(5-4)/6+(10-9)/6=2/6=1/3$
Now I am asking the following:
Let $X_1, X_2, X_3, dots, X_{16}$ be a random sample from this uniform distribution. We are asked to find Chebyshev's upper bound for $P(overline{X} leq 5 text { or } overline{X} geq 9)$ where $overline{X}=dfrac{sum_{i=1}^{16}X_i}{16}$
Again, we can break up the probability to be as follows:
$P(overline{X} leq 5) + P(overline{X} geq 9)leq text{upper bound}$
Chebyshev Inequality is the following: $P(|X-mu_x| geq a)leq dfrac{var(x)}{a^2}$
So, for $P(overline{X}leq 5)$, we have $P(|overline{X}-7|leq5)=1-dfrac{3}{25}$
For $P(overline{X}geq 9)$, we have $P(|overline{X}-7|geq 9)=dfrac{3}{9^2}$
So in total, the upper bound must be $1-dfrac{3}{25}+dfrac{3}{9^2}=0.917$
Is this correct? I don't see where we used $overline{X}$ in this. There would be nothing different if I considered $X$, instead of $overline{X}$, so I think I am missing something.
Maybe the $var(x)$ should actually be $sigma^2/n=3/16$. But in my notes, I have that this is only the case if $X_1, X_2, dots, X_n$ are $N(mu, sigma^2)$. Here we have that they are uniform, so I am not sure.
probability statistics inequality
probability statistics inequality
asked Dec 5 '18 at 21:17
K Split XK Split X
4,19611131
asked Dec 5 '18 at 21:17
K Split XK Split X
4,19611131
asked Dec 5 '18 at 21:17
K Split XK Split X
4,19611131
asked Dec 5 '18 at 21:17
K Split XK Split X
4,19611131
asked Dec 5 '18 at 21:17
K Split XK Split X
4,19611131
4,19611131
$begingroup$
In general, $text{var}(bar{X}) = frac{1}{n^2} text{var}(sum_{i=1}^n X_i) = frac{1}{n} text{var}(X_1)$ for i.i.d. $X_i$. You should use this variance when applying Chebychev to $bar{X}$.
$endgroup$
– angryavian
Dec 5 '18 at 21:27
add a comment |
$begingroup$
In general, $text{var}(bar{X}) = frac{1}{n^2} text{var}(sum_{i=1}^n X_i) = frac{1}{n} text{var}(X_1)$ for i.i.d. $X_i$. You should use this variance when applying Chebychev to $bar{X}$.
$endgroup$
– angryavian
Dec 5 '18 at 21:27
$begingroup$
In general, $text{var}(bar{X}) = frac{1}{n^2} text{var}(sum_{i=1}^n X_i) = frac{1}{n} text{var}(X_1)$ for i.i.d. $X_i$. You should use this variance when applying Chebychev to $bar{X}$.
$endgroup$
– angryavian
Dec 5 '18 at 21:27
$begingroup$
In general, $text{var}(bar{X}) = frac{1}{n^2} text{var}(sum_{i=1}^n X_i) = frac{1}{n} text{var}(X_1)$ for i.i.d. $X_i$. You should use this variance when applying Chebychev to $bar{X}$.
$endgroup$
– angryavian
Dec 5 '18 at 21:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
$P(X leq 5 text { or } X geq 9) = P(|X-7| geq 2) $
two away from mean on both sides..
answered Dec 5 '18 at 21:57
karakfakarakfa
1,973811
$endgroup$
$begingroup$
This would give a different upper bound then what I got
$endgroup$
– K Split X
Dec 5 '18 at 22:40
$begingroup$
Yes, because your calculations steps are not correct. $P(overline{X }leq 5)$, doesn't mean $P(|overline{X}-7| leq 5)$ Just put $X=1$ to check. Your answer is also counterintuive.
$endgroup$
– karakfa
Dec 5 '18 at 22:45
$begingroup$
What is the correct way to solve the $P(overline{X} lt 5)$ ?
$endgroup$
– K Split X
Dec 5 '18 at 22:50
$begingroup$
Follow my hint, you can solve both sides at the same time. This is a direct application of Chebyshev Inequality, nothing else needed.
$endgroup$
– karakfa
Dec 5 '18 at 22:53
$begingroup$
If instead, the mean was something like $8.5$, how would we solve this? In this question, the mean is in the center of $5$ and $9$, but what if its not?
$endgroup$
– K Split X
Dec 6 '18 at 0:27
|
show 5 more comments
Your Answer
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1 Answer
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1 Answer
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oldest
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$begingroup$
Hint:
$P(X leq 5 text { or } X geq 9) = P(|X-7| geq 2) $
two away from mean on both sides..
answered Dec 5 '18 at 21:57
karakfakarakfa
1,973811
$endgroup$
$begingroup$
This would give a different upper bound then what I got
$endgroup$
– K Split X
Dec 5 '18 at 22:40
$begingroup$
Yes, because your calculations steps are not correct. $P(overline{X }leq 5)$, doesn't mean $P(|overline{X}-7| leq 5)$ Just put $X=1$ to check. Your answer is also counterintuive.
$endgroup$
– karakfa
Dec 5 '18 at 22:45
$begingroup$
What is the correct way to solve the $P(overline{X} lt 5)$ ?
$endgroup$
– K Split X
Dec 5 '18 at 22:50
$begingroup$
Follow my hint, you can solve both sides at the same time. This is a direct application of Chebyshev Inequality, nothing else needed.
$endgroup$
– karakfa
Dec 5 '18 at 22:53
$begingroup$
If instead, the mean was something like $8.5$, how would we solve this? In this question, the mean is in the center of $5$ and $9$, but what if its not?
$endgroup$
– K Split X
Dec 6 '18 at 0:27
|
show 5 more comments
$begingroup$
Hint:
$P(X leq 5 text { or } X geq 9) = P(|X-7| geq 2) $
two away from mean on both sides..
answered Dec 5 '18 at 21:57
karakfakarakfa
1,973811
$endgroup$
$begingroup$
This would give a different upper bound then what I got
$endgroup$
– K Split X
Dec 5 '18 at 22:40
$begingroup$
Yes, because your calculations steps are not correct. $P(overline{X }leq 5)$, doesn't mean $P(|overline{X}-7| leq 5)$ Just put $X=1$ to check. Your answer is also counterintuive.
$endgroup$
– karakfa
Dec 5 '18 at 22:45
$begingroup$
What is the correct way to solve the $P(overline{X} lt 5)$ ?
$endgroup$
– K Split X
Dec 5 '18 at 22:50
$begingroup$
Follow my hint, you can solve both sides at the same time. This is a direct application of Chebyshev Inequality, nothing else needed.
$endgroup$
– karakfa
Dec 5 '18 at 22:53
$begingroup$
If instead, the mean was something like $8.5$, how would we solve this? In this question, the mean is in the center of $5$ and $9$, but what if its not?
$endgroup$
– K Split X
Dec 6 '18 at 0:27
|
show 5 more comments
$begingroup$
Hint:
$P(X leq 5 text { or } X geq 9) = P(|X-7| geq 2) $
two away from mean on both sides..
answered Dec 5 '18 at 21:57
karakfakarakfa
1,973811
$endgroup$
Hint:
$P(X leq 5 text { or } X geq 9) = P(|X-7| geq 2) $
two away from mean on both sides..
answered Dec 5 '18 at 21:57
karakfakarakfa
1,973811
answered Dec 5 '18 at 21:57
karakfakarakfa
1,973811
answered Dec 5 '18 at 21:57
karakfakarakfa
1,973811
answered Dec 5 '18 at 21:57
karakfakarakfa
1,973811
1,973811
$begingroup$
This would give a different upper bound then what I got
$endgroup$
– K Split X
Dec 5 '18 at 22:40
$begingroup$
Yes, because your calculations steps are not correct. $P(overline{X }leq 5)$, doesn't mean $P(|overline{X}-7| leq 5)$ Just put $X=1$ to check. Your answer is also counterintuive.
$endgroup$
– karakfa
Dec 5 '18 at 22:45
$begingroup$
What is the correct way to solve the $P(overline{X} lt 5)$ ?
$endgroup$
– K Split X
Dec 5 '18 at 22:50
$begingroup$
Follow my hint, you can solve both sides at the same time. This is a direct application of Chebyshev Inequality, nothing else needed.
$endgroup$
– karakfa
Dec 5 '18 at 22:53
$begingroup$
If instead, the mean was something like $8.5$, how would we solve this? In this question, the mean is in the center of $5$ and $9$, but what if its not?
$endgroup$
– K Split X
Dec 6 '18 at 0:27
|
show 5 more comments
$begingroup$
This would give a different upper bound then what I got
$endgroup$
– K Split X
Dec 5 '18 at 22:40
$begingroup$
Yes, because your calculations steps are not correct. $P(overline{X }leq 5)$, doesn't mean $P(|overline{X}-7| leq 5)$ Just put $X=1$ to check. Your answer is also counterintuive.
$endgroup$
– karakfa
Dec 5 '18 at 22:45
$begingroup$
What is the correct way to solve the $P(overline{X} lt 5)$ ?
$endgroup$
– K Split X
Dec 5 '18 at 22:50
$begingroup$
Follow my hint, you can solve both sides at the same time. This is a direct application of Chebyshev Inequality, nothing else needed.
$endgroup$
– karakfa
Dec 5 '18 at 22:53
$begingroup$
If instead, the mean was something like $8.5$, how would we solve this? In this question, the mean is in the center of $5$ and $9$, but what if its not?
$endgroup$
– K Split X
Dec 6 '18 at 0:27
$begingroup$
This would give a different upper bound then what I got
$endgroup$
– K Split X
Dec 5 '18 at 22:40
$begingroup$
This would give a different upper bound then what I got
$endgroup$
– K Split X
Dec 5 '18 at 22:40
$begingroup$
Yes, because your calculations steps are not correct. $P(overline{X }leq 5)$, doesn't mean $P(|overline{X}-7| leq 5)$ Just put $X=1$ to check. Your answer is also counterintuive.
$endgroup$
– karakfa
Dec 5 '18 at 22:45
$begingroup$
Yes, because your calculations steps are not correct. $P(overline{X }leq 5)$, doesn't mean $P(|overline{X}-7| leq 5)$ Just put $X=1$ to check. Your answer is also counterintuive.
$endgroup$
– karakfa
Dec 5 '18 at 22:45
$begingroup$
What is the correct way to solve the $P(overline{X} lt 5)$ ?
$endgroup$
– K Split X
Dec 5 '18 at 22:50
$begingroup$
What is the correct way to solve the $P(overline{X} lt 5)$ ?
$endgroup$
– K Split X
Dec 5 '18 at 22:50
$begingroup$
Follow my hint, you can solve both sides at the same time. This is a direct application of Chebyshev Inequality, nothing else needed.
$endgroup$
– karakfa
Dec 5 '18 at 22:53
$begingroup$
Follow my hint, you can solve both sides at the same time. This is a direct application of Chebyshev Inequality, nothing else needed.
$endgroup$
– karakfa
Dec 5 '18 at 22:53
$begingroup$
If instead, the mean was something like $8.5$, how would we solve this? In this question, the mean is in the center of $5$ and $9$, but what if its not?
$endgroup$
– K Split X
Dec 6 '18 at 0:27
$begingroup$
If instead, the mean was something like $8.5$, how would we solve this? In this question, the mean is in the center of $5$ and $9$, but what if its not?
$endgroup$
– K Split X
Dec 6 '18 at 0:27
|
show 5 more comments
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$begingroup$
In general, $text{var}(bar{X}) = frac{1}{n^2} text{var}(sum_{i=1}^n X_i) = frac{1}{n} text{var}(X_1)$ for i.i.d. $X_i$. You should use this variance when applying Chebychev to $bar{X}$.
$endgroup$
– angryavian
Dec 5 '18 at 21:27