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 at 18:07
agblt
13513
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 at 18:07
agblt
13513
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 at 18:07
agblt
13513
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 at 18:07
agblt
13513
asked Nov 28 at 18:07
agblt
13513
asked Nov 28 at 18:07
agblt
13513
asked Nov 28 at 18:07
agblt
13513
asked Nov 28 at 18:07
agblt
13513
13513
add a comment |
add a comment |
1 Answer
1
active
oldest
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 at 18:16
Robert Israel
317k23206457
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017486%2ftrouble-understanding-when-to-use-operatornamevarax-a2-operatornamevar%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
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 at 18:16
Robert Israel
317k23206457
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 at 18:16
Robert Israel
317k23206457
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 at 18:16
Robert Israel
317k23206457
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 at 18:16
Robert Israel
317k23206457
answered Nov 28 at 18:16
Robert Israel
317k23206457
answered Nov 28 at 18:16
Robert Israel
317k23206457
answered Nov 28 at 18:16
Robert Israel
317k23206457
317k23206457
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017486%2ftrouble-understanding-when-to-use-operatornamevarax-a2-operatornamevar%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
