Trouble understanding when to use $operatorname{Var}(aX) = a^2operatorname{Var}(X)$
From SOA Sample #203
A machine has two components and fails when both components fail. The number of
years from now until the first component fails, X, and the number of years from now until
the machine fails, Y, are random variables with joint density function
$$f(x,y)=begin{cases}frac1{18}e^{-(x+y)over6}&text{if }0<x<y\
0&text{otherwise}end{cases}$$
Calculate $operatorname{Var}(Y|X=2)$.
To begin one must calculate $f_{Y|X}(y|2)$. I eventually get
$$ {{{1over18}e^{-(2+y)over6}}over{{1over3}e^{-{4over 6}}}}={{1over6}e^{2-yover6}}$$
At this point I can factor out the $e^{2over6}$ and will be left with the exponential distribution times a constant:
$$e^{2over6}{{1over6}e^{-yover6}}$$
I thought that one should apply the rule $operatorname{Var}(aX) = a^2operatorname{Var}(X)$ here.
$$operatorname{Var}(e^{2over6}Y)=e^{4over6}operatorname{Var}(Y)=e^{4over6}cdot36$$
However the true solution is $36$. Why am I wrong?
probability statistics
asked Nov 28 '18 at 18:07
agbltagblt
17414
add a comment |
From SOA Sample #203
A machine has two components and fails when both components fail. The number of
years from now until the first component fails, X, and the number of years from now until
the machine fails, Y, are random variables with joint density function
$$f(x,y)=begin{cases}frac1{18}e^{-(x+y)over6}&text{if }0<x<y\
0&text{otherwise}end{cases}$$
Calculate $operatorname{Var}(Y|X=2)$.
To begin one must calculate $f_{Y|X}(y|2)$. I eventually get
$$ {{{1over18}e^{-(2+y)over6}}over{{1over3}e^{-{4over 6}}}}={{1over6}e^{2-yover6}}$$
At this point I can factor out the $e^{2over6}$ and will be left with the exponential distribution times a constant:
$$e^{2over6}{{1over6}e^{-yover6}}$$
I thought that one should apply the rule $operatorname{Var}(aX) = a^2operatorname{Var}(X)$ here.
$$operatorname{Var}(e^{2over6}Y)=e^{4over6}operatorname{Var}(Y)=e^{4over6}cdot36$$
However the true solution is $36$. Why am I wrong?
probability statistics
asked Nov 28 '18 at 18:07
agbltagblt
17414
add a comment |
From SOA Sample #203
A machine has two components and fails when both components fail. The number of
years from now until the first component fails, X, and the number of years from now until
the machine fails, Y, are random variables with joint density function
$$f(x,y)=begin{cases}frac1{18}e^{-(x+y)over6}&text{if }0<x<y\
0&text{otherwise}end{cases}$$
Calculate $operatorname{Var}(Y|X=2)$.
To begin one must calculate $f_{Y|X}(y|2)$. I eventually get
$$ {{{1over18}e^{-(2+y)over6}}over{{1over3}e^{-{4over 6}}}}={{1over6}e^{2-yover6}}$$
At this point I can factor out the $e^{2over6}$ and will be left with the exponential distribution times a constant:
$$e^{2over6}{{1over6}e^{-yover6}}$$
I thought that one should apply the rule $operatorname{Var}(aX) = a^2operatorname{Var}(X)$ here.
$$operatorname{Var}(e^{2over6}Y)=e^{4over6}operatorname{Var}(Y)=e^{4over6}cdot36$$
However the true solution is $36$. Why am I wrong?
probability statistics
asked Nov 28 '18 at 18:07
agbltagblt
17414
From SOA Sample #203
A machine has two components and fails when both components fail. The number of
years from now until the first component fails, X, and the number of years from now until
the machine fails, Y, are random variables with joint density function
$$f(x,y)=begin{cases}frac1{18}e^{-(x+y)over6}&text{if }0<x<y\
0&text{otherwise}end{cases}$$
Calculate $operatorname{Var}(Y|X=2)$.
To begin one must calculate $f_{Y|X}(y|2)$. I eventually get
$$ {{{1over18}e^{-(2+y)over6}}over{{1over3}e^{-{4over 6}}}}={{1over6}e^{2-yover6}}$$
At this point I can factor out the $e^{2over6}$ and will be left with the exponential distribution times a constant:
$$e^{2over6}{{1over6}e^{-yover6}}$$
I thought that one should apply the rule $operatorname{Var}(aX) = a^2operatorname{Var}(X)$ here.
$$operatorname{Var}(e^{2over6}Y)=e^{4over6}operatorname{Var}(Y)=e^{4over6}cdot36$$
However the true solution is $36$. Why am I wrong?
probability statistics
probability statistics
asked Nov 28 '18 at 18:07
agbltagblt
17414
asked Nov 28 '18 at 18:07
agbltagblt
17414
asked Nov 28 '18 at 18:07
agbltagblt
17414
asked Nov 28 '18 at 18:07
agbltagblt
17414
asked Nov 28 '18 at 18:07
agbltagblt
17414
17414
add a comment |
add a comment |
1 Answer
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This is not "exponential distribution times a constant", it is a shifted exponential distribution. Note that your formula for the conditional density is only valid for $y > 2$. Thus the conditional distribution for $Y$ given $X = 2$ is the distribution of $2 + T$ where $T$ is exponential with rate $1/6$. The variance of $2+T$ is the same as the variance of $T$.
answered Nov 28 '18 at 18:16
Robert IsraelRobert Israel
319k23208457
add a comment |
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1 Answer
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1 Answer
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active
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votes
This is not "exponential distribution times a constant", it is a shifted exponential distribution. Note that your formula for the conditional density is only valid for $y > 2$. Thus the conditional distribution for $Y$ given $X = 2$ is the distribution of $2 + T$ where $T$ is exponential with rate $1/6$. The variance of $2+T$ is the same as the variance of $T$.
answered Nov 28 '18 at 18:16
Robert IsraelRobert Israel
319k23208457
add a comment |
This is not "exponential distribution times a constant", it is a shifted exponential distribution. Note that your formula for the conditional density is only valid for $y > 2$. Thus the conditional distribution for $Y$ given $X = 2$ is the distribution of $2 + T$ where $T$ is exponential with rate $1/6$. The variance of $2+T$ is the same as the variance of $T$.
answered Nov 28 '18 at 18:16
Robert IsraelRobert Israel
319k23208457
add a comment |
This is not "exponential distribution times a constant", it is a shifted exponential distribution. Note that your formula for the conditional density is only valid for $y > 2$. Thus the conditional distribution for $Y$ given $X = 2$ is the distribution of $2 + T$ where $T$ is exponential with rate $1/6$. The variance of $2+T$ is the same as the variance of $T$.
answered Nov 28 '18 at 18:16
Robert IsraelRobert Israel
319k23208457
This is not "exponential distribution times a constant", it is a shifted exponential distribution. Note that your formula for the conditional density is only valid for $y > 2$. Thus the conditional distribution for $Y$ given $X = 2$ is the distribution of $2 + T$ where $T$ is exponential with rate $1/6$. The variance of $2+T$ is the same as the variance of $T$.
answered Nov 28 '18 at 18:16
Robert IsraelRobert Israel
319k23208457
answered Nov 28 '18 at 18:16
Robert IsraelRobert Israel
319k23208457
answered Nov 28 '18 at 18:16
Robert IsraelRobert Israel
319k23208457
answered Nov 28 '18 at 18:16
Robert IsraelRobert Israel
319k23208457
319k23208457
add a comment |
add a comment |
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