what is the expectation of this random variable?
Suppose X is an Hyper Geometric random variable with parameters w,b and n, that is, X ~ HGeom(w,b,n).
And suppose Y = $Xchoose2$
, what is E(Y)? Using the sum of probability times the value of Y yields a very complicated formula, there should be some clever way to get the result. Could some one help with this?
Thanks.
probability expected-value
asked Dec 4 '18 at 0:53
wangshuaijiewangshuaijie
1638
add a comment |
Suppose X is an Hyper Geometric random variable with parameters w,b and n, that is, X ~ HGeom(w,b,n).
And suppose Y = $Xchoose2$
, what is E(Y)? Using the sum of probability times the value of Y yields a very complicated formula, there should be some clever way to get the result. Could some one help with this?
Thanks.
probability expected-value
asked Dec 4 '18 at 0:53
wangshuaijiewangshuaijie
1638
add a comment |
Suppose X is an Hyper Geometric random variable with parameters w,b and n, that is, X ~ HGeom(w,b,n).
And suppose Y = $Xchoose2$
, what is E(Y)? Using the sum of probability times the value of Y yields a very complicated formula, there should be some clever way to get the result. Could some one help with this?
Thanks.
probability expected-value
asked Dec 4 '18 at 0:53
wangshuaijiewangshuaijie
1638
Suppose X is an Hyper Geometric random variable with parameters w,b and n, that is, X ~ HGeom(w,b,n).
And suppose Y = $Xchoose2$
, what is E(Y)? Using the sum of probability times the value of Y yields a very complicated formula, there should be some clever way to get the result. Could some one help with this?
Thanks.
probability expected-value
probability expected-value
asked Dec 4 '18 at 0:53
wangshuaijiewangshuaijie
1638
asked Dec 4 '18 at 0:53
wangshuaijiewangshuaijie
1638
asked Dec 4 '18 at 0:53
wangshuaijiewangshuaijie
1638
asked Dec 4 '18 at 0:53
wangshuaijiewangshuaijie
1638
asked Dec 4 '18 at 0:53
wangshuaijiewangshuaijie
1638
1638
add a comment |
add a comment |
2 Answers
2
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votes
First, $Y={ X choose 2}=frac{X!}{2!(X-2)!}=frac{X(X-1)(X-2)!}{2!(X-2)!}=frac{X(X-1)}{2}$.
Second, use the fact that for any random variable $X$ and $a, b in mathbb{R}$, $E(aX+b)=aE(X)+b$.
Thirdly, $Var(X)=E(X^{2})-E(X)^{2}$.
So, $E(Y)=E(frac{X(X-1)}{2})=E(frac{X^{2}-X}{2})=frac{E(X^{2})}{2}-frac{E(X)}{2}$
All that is left is for you to find the expectation and variance of $X$ (so that you can find $E(X^{2})$). You can find the formulas in any textbook or online (Wikipedia).
answered Dec 4 '18 at 2:31
Live Free or π HardLive Free or π Hard
479213
add a comment |
Guide:
The mean and variace of Hypergeometric should be well known, that is we know the first two moments.
Express $Y$ as a linear function of $X$ and $X^2$ and use linearity of expectation.
answered Dec 4 '18 at 0:57
Siong Thye GohSiong Thye Goh
99.9k1465117
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
First, $Y={ X choose 2}=frac{X!}{2!(X-2)!}=frac{X(X-1)(X-2)!}{2!(X-2)!}=frac{X(X-1)}{2}$.
Second, use the fact that for any random variable $X$ and $a, b in mathbb{R}$, $E(aX+b)=aE(X)+b$.
Thirdly, $Var(X)=E(X^{2})-E(X)^{2}$.
So, $E(Y)=E(frac{X(X-1)}{2})=E(frac{X^{2}-X}{2})=frac{E(X^{2})}{2}-frac{E(X)}{2}$
All that is left is for you to find the expectation and variance of $X$ (so that you can find $E(X^{2})$). You can find the formulas in any textbook or online (Wikipedia).
answered Dec 4 '18 at 2:31
Live Free or π HardLive Free or π Hard
479213
add a comment |
First, $Y={ X choose 2}=frac{X!}{2!(X-2)!}=frac{X(X-1)(X-2)!}{2!(X-2)!}=frac{X(X-1)}{2}$.
Second, use the fact that for any random variable $X$ and $a, b in mathbb{R}$, $E(aX+b)=aE(X)+b$.
Thirdly, $Var(X)=E(X^{2})-E(X)^{2}$.
So, $E(Y)=E(frac{X(X-1)}{2})=E(frac{X^{2}-X}{2})=frac{E(X^{2})}{2}-frac{E(X)}{2}$
All that is left is for you to find the expectation and variance of $X$ (so that you can find $E(X^{2})$). You can find the formulas in any textbook or online (Wikipedia).
answered Dec 4 '18 at 2:31
Live Free or π HardLive Free or π Hard
479213
add a comment |
First, $Y={ X choose 2}=frac{X!}{2!(X-2)!}=frac{X(X-1)(X-2)!}{2!(X-2)!}=frac{X(X-1)}{2}$.
Second, use the fact that for any random variable $X$ and $a, b in mathbb{R}$, $E(aX+b)=aE(X)+b$.
Thirdly, $Var(X)=E(X^{2})-E(X)^{2}$.
So, $E(Y)=E(frac{X(X-1)}{2})=E(frac{X^{2}-X}{2})=frac{E(X^{2})}{2}-frac{E(X)}{2}$
All that is left is for you to find the expectation and variance of $X$ (so that you can find $E(X^{2})$). You can find the formulas in any textbook or online (Wikipedia).
answered Dec 4 '18 at 2:31
Live Free or π HardLive Free or π Hard
479213
First, $Y={ X choose 2}=frac{X!}{2!(X-2)!}=frac{X(X-1)(X-2)!}{2!(X-2)!}=frac{X(X-1)}{2}$.
Second, use the fact that for any random variable $X$ and $a, b in mathbb{R}$, $E(aX+b)=aE(X)+b$.
Thirdly, $Var(X)=E(X^{2})-E(X)^{2}$.
So, $E(Y)=E(frac{X(X-1)}{2})=E(frac{X^{2}-X}{2})=frac{E(X^{2})}{2}-frac{E(X)}{2}$
All that is left is for you to find the expectation and variance of $X$ (so that you can find $E(X^{2})$). You can find the formulas in any textbook or online (Wikipedia).
answered Dec 4 '18 at 2:31
Live Free or π HardLive Free or π Hard
479213
edited Dec 4 '18 at 2:37
answered Dec 4 '18 at 2:31
Live Free or π HardLive Free or π Hard
479213
answered Dec 4 '18 at 2:31
Live Free or π HardLive Free or π Hard
479213
answered Dec 4 '18 at 2:31
Live Free or π HardLive Free or π Hard
479213
479213
add a comment |
add a comment |
Guide:
The mean and variace of Hypergeometric should be well known, that is we know the first two moments.
Express $Y$ as a linear function of $X$ and $X^2$ and use linearity of expectation.
answered Dec 4 '18 at 0:57
Siong Thye GohSiong Thye Goh
99.9k1465117
add a comment |
Guide:
The mean and variace of Hypergeometric should be well known, that is we know the first two moments.
Express $Y$ as a linear function of $X$ and $X^2$ and use linearity of expectation.
answered Dec 4 '18 at 0:57
Siong Thye GohSiong Thye Goh
99.9k1465117
add a comment |
Guide:
The mean and variace of Hypergeometric should be well known, that is we know the first two moments.
Express $Y$ as a linear function of $X$ and $X^2$ and use linearity of expectation.
answered Dec 4 '18 at 0:57
Siong Thye GohSiong Thye Goh
99.9k1465117
Guide:
The mean and variace of Hypergeometric should be well known, that is we know the first two moments.
Express $Y$ as a linear function of $X$ and $X^2$ and use linearity of expectation.
answered Dec 4 '18 at 0:57
Siong Thye GohSiong Thye Goh
99.9k1465117
answered Dec 4 '18 at 0:57
Siong Thye GohSiong Thye Goh
99.9k1465117
answered Dec 4 '18 at 0:57
Siong Thye GohSiong Thye Goh
99.9k1465117
answered Dec 4 '18 at 0:57
Siong Thye GohSiong Thye Goh
99.9k1465117
99.9k1465117
add a comment |
add a comment |
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