Mixing
Transformation of random variables: what went wrong?
$begingroup$
Assume that $X_{1}$ and $X_{2}$ are independent exponential random variables with parameter $lambda$. Let $Y_{1} = X_{1} + X_{2}$ and $Y_{2} = X_{1} - X_{2}$. Determine
(a) Find the joint distribution of $Y_{1}$ and $Y_{2}$
(b) Find the marginal distribution of $Y_{1}$
(c) Find the marginal distribution of $Y_{2}$
d) Are $Y_{1}$ and $Y_{2}$ independent?
MY SOLUTION
(a) To begin with, I noticed that $X_{1} = (Y_{1} + Y_{2})/2$ and $X_{2} = (Y_{1} - Y_{2})/2$. Hence we get
begin{align*}
f_{Y_{1},Y_{2}}(Y_{1},Y_{2}) & = f_{X_{1},X_{2}}left(frac{Y_{1}+Y_{2}}{2},frac{Y_{1}-Y_{2}}{2}right)|det J(y_{1},y_{2})|\\
& = frac{1}{2}f_{X_{1}}left(frac{Y_{1} + Y_{2}}{2}right)f_{X_{2}}left(frac{Y_{1}-Y_{2}}{2}right) = frac{lambda^{2}}{2}exp(-lambda y_{1})
end{align*}
where the support of $Y$ is given by the intersection of $Y_{1} - Y_{2} geq 0$ and $Y_{1} + Y_{2} geq 0$.
(b) According to the definition of marginal distribution, we have
begin{align*}
f_{Y_{1}}(y_{1}) = int_{-infty}^{+infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})mathrm{d}y_{2} = int_{-y_{1}}^{y_{1}}frac{lambda^{2}}{2}exp(-lambda y_{1})mathrm{d}y_{2} = lambda^{2}y_{1}exp(-lambda y_{1})
end{align*}
where $y_{1} geq 0$.
(c) Analogously, we have
begin{align*}
f_{Y_{2}}(y_{2}) = int_{-infty}^{+infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})mathrm{d}y_{1} = int_{y_{2}}^{infty}frac{lambda^{2}}{2}exp(-lambda y_{1})mathrm{d}y_{1} = frac{lambda}{2}exp(-lambda y_{2})
end{align*}
Then I get stuck, because I do not know how to describe the support of $Y_{2}$ neither if the integration limits are wrong. Could someone help me out?
probability-theory proof-verification probability-distributions change-of-variable
edited Feb 1 at 20:13
Did
249k23228466
asked Feb 1 at 19:48
APC89APC89
2,371720
$endgroup$
|
show 1 more comment
$begingroup$
Assume that $X_{1}$ and $X_{2}$ are independent exponential random variables with parameter $lambda$. Let $Y_{1} = X_{1} + X_{2}$ and $Y_{2} = X_{1} - X_{2}$. Determine
(a) Find the joint distribution of $Y_{1}$ and $Y_{2}$
(b) Find the marginal distribution of $Y_{1}$
(c) Find the marginal distribution of $Y_{2}$
d) Are $Y_{1}$ and $Y_{2}$ independent?
MY SOLUTION
(a) To begin with, I noticed that $X_{1} = (Y_{1} + Y_{2})/2$ and $X_{2} = (Y_{1} - Y_{2})/2$. Hence we get
begin{align*}
f_{Y_{1},Y_{2}}(Y_{1},Y_{2}) & = f_{X_{1},X_{2}}left(frac{Y_{1}+Y_{2}}{2},frac{Y_{1}-Y_{2}}{2}right)|det J(y_{1},y_{2})|\\
& = frac{1}{2}f_{X_{1}}left(frac{Y_{1} + Y_{2}}{2}right)f_{X_{2}}left(frac{Y_{1}-Y_{2}}{2}right) = frac{lambda^{2}}{2}exp(-lambda y_{1})
end{align*}
where the support of $Y$ is given by the intersection of $Y_{1} - Y_{2} geq 0$ and $Y_{1} + Y_{2} geq 0$.
(b) According to the definition of marginal distribution, we have
begin{align*}
f_{Y_{1}}(y_{1}) = int_{-infty}^{+infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})mathrm{d}y_{2} = int_{-y_{1}}^{y_{1}}frac{lambda^{2}}{2}exp(-lambda y_{1})mathrm{d}y_{2} = lambda^{2}y_{1}exp(-lambda y_{1})
end{align*}
where $y_{1} geq 0$.
(c) Analogously, we have
begin{align*}
f_{Y_{2}}(y_{2}) = int_{-infty}^{+infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})mathrm{d}y_{1} = int_{y_{2}}^{infty}frac{lambda^{2}}{2}exp(-lambda y_{1})mathrm{d}y_{1} = frac{lambda}{2}exp(-lambda y_{2})
end{align*}
Then I get stuck, because I do not know how to describe the support of $Y_{2}$ neither if the integration limits are wrong. Could someone help me out?
probability-theory proof-verification probability-distributions change-of-variable
edited Feb 1 at 20:13
Did
249k23228466
asked Feb 1 at 19:48
APC89APC89
2,371720
$endgroup$
$begingroup$
the support of $Y_2 + X_2$ is $mathbb{R^{+}}$ and the support of $X_2$ is $mathbb{R^{+}}$, from this you can conclude that the support of $Y_2$ is $mathbb{R}$
$endgroup$
– rapidracim
Feb 1 at 20:02
$begingroup$
I thought so, but the problem is the marginal probability density function expression: when you make $y_{2}$ varies from $-infty$ to $+infty$, the integral diverges. So could you help me find out where is the mistake? By the way, if the one half disappears, we get an exponential random variable.
$endgroup$
– APC89
Feb 1 at 20:04
$begingroup$
btw have you tried finding the distribution of $-X_2$ and then use the convolution formula for the distribution of $Y_2$ ?
$endgroup$
– rapidracim
Feb 1 at 20:09
$begingroup$
We haven't studied the convolution formula yet.
$endgroup$
– APC89
Feb 1 at 20:10
1
$begingroup$
The solution of (c) should read $$f_{Y_2}(y_2) = int_{|y_2|}^infty frac{lambda^2}2exp(-lambda y_1)mathrm dy_1 = frac{lambda}2exp(-lambda |y_2|) $$ for every $y_2$ in $mathbb R$.
$endgroup$
– Did
Feb 1 at 20:13
|
show 1 more comment
$begingroup$
Assume that $X_{1}$ and $X_{2}$ are independent exponential random variables with parameter $lambda$. Let $Y_{1} = X_{1} + X_{2}$ and $Y_{2} = X_{1} - X_{2}$. Determine
(a) Find the joint distribution of $Y_{1}$ and $Y_{2}$
(b) Find the marginal distribution of $Y_{1}$
(c) Find the marginal distribution of $Y_{2}$
d) Are $Y_{1}$ and $Y_{2}$ independent?
MY SOLUTION
(a) To begin with, I noticed that $X_{1} = (Y_{1} + Y_{2})/2$ and $X_{2} = (Y_{1} - Y_{2})/2$. Hence we get
begin{align*}
f_{Y_{1},Y_{2}}(Y_{1},Y_{2}) & = f_{X_{1},X_{2}}left(frac{Y_{1}+Y_{2}}{2},frac{Y_{1}-Y_{2}}{2}right)|det J(y_{1},y_{2})|\\
& = frac{1}{2}f_{X_{1}}left(frac{Y_{1} + Y_{2}}{2}right)f_{X_{2}}left(frac{Y_{1}-Y_{2}}{2}right) = frac{lambda^{2}}{2}exp(-lambda y_{1})
end{align*}
where the support of $Y$ is given by the intersection of $Y_{1} - Y_{2} geq 0$ and $Y_{1} + Y_{2} geq 0$.
(b) According to the definition of marginal distribution, we have
begin{align*}
f_{Y_{1}}(y_{1}) = int_{-infty}^{+infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})mathrm{d}y_{2} = int_{-y_{1}}^{y_{1}}frac{lambda^{2}}{2}exp(-lambda y_{1})mathrm{d}y_{2} = lambda^{2}y_{1}exp(-lambda y_{1})
end{align*}
where $y_{1} geq 0$.
(c) Analogously, we have
begin{align*}
f_{Y_{2}}(y_{2}) = int_{-infty}^{+infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})mathrm{d}y_{1} = int_{y_{2}}^{infty}frac{lambda^{2}}{2}exp(-lambda y_{1})mathrm{d}y_{1} = frac{lambda}{2}exp(-lambda y_{2})
end{align*}
Then I get stuck, because I do not know how to describe the support of $Y_{2}$ neither if the integration limits are wrong. Could someone help me out?
probability-theory proof-verification probability-distributions change-of-variable
edited Feb 1 at 20:13
Did
249k23228466
asked Feb 1 at 19:48
APC89APC89
2,371720
$endgroup$
Assume that $X_{1}$ and $X_{2}$ are independent exponential random variables with parameter $lambda$. Let $Y_{1} = X_{1} + X_{2}$ and $Y_{2} = X_{1} - X_{2}$. Determine
(a) Find the joint distribution of $Y_{1}$ and $Y_{2}$
(b) Find the marginal distribution of $Y_{1}$
(c) Find the marginal distribution of $Y_{2}$
d) Are $Y_{1}$ and $Y_{2}$ independent?
MY SOLUTION
(a) To begin with, I noticed that $X_{1} = (Y_{1} + Y_{2})/2$ and $X_{2} = (Y_{1} - Y_{2})/2$. Hence we get
begin{align*}
f_{Y_{1},Y_{2}}(Y_{1},Y_{2}) & = f_{X_{1},X_{2}}left(frac{Y_{1}+Y_{2}}{2},frac{Y_{1}-Y_{2}}{2}right)|det J(y_{1},y_{2})|\\
& = frac{1}{2}f_{X_{1}}left(frac{Y_{1} + Y_{2}}{2}right)f_{X_{2}}left(frac{Y_{1}-Y_{2}}{2}right) = frac{lambda^{2}}{2}exp(-lambda y_{1})
end{align*}
where the support of $Y$ is given by the intersection of $Y_{1} - Y_{2} geq 0$ and $Y_{1} + Y_{2} geq 0$.
(b) According to the definition of marginal distribution, we have
begin{align*}
f_{Y_{1}}(y_{1}) = int_{-infty}^{+infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})mathrm{d}y_{2} = int_{-y_{1}}^{y_{1}}frac{lambda^{2}}{2}exp(-lambda y_{1})mathrm{d}y_{2} = lambda^{2}y_{1}exp(-lambda y_{1})
end{align*}
where $y_{1} geq 0$.
(c) Analogously, we have
begin{align*}
f_{Y_{2}}(y_{2}) = int_{-infty}^{+infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})mathrm{d}y_{1} = int_{y_{2}}^{infty}frac{lambda^{2}}{2}exp(-lambda y_{1})mathrm{d}y_{1} = frac{lambda}{2}exp(-lambda y_{2})
end{align*}
Then I get stuck, because I do not know how to describe the support of $Y_{2}$ neither if the integration limits are wrong. Could someone help me out?
probability-theory proof-verification probability-distributions change-of-variable
probability-theory proof-verification probability-distributions change-of-variable
edited Feb 1 at 20:13
Did
249k23228466
asked Feb 1 at 19:48
APC89APC89
2,371720
edited Feb 1 at 20:13
Did
249k23228466
asked Feb 1 at 19:48
APC89APC89
2,371720
edited Feb 1 at 20:13
Did
249k23228466
edited Feb 1 at 20:13
Did
249k23228466
edited Feb 1 at 20:13
Did
249k23228466
249k23228466
asked Feb 1 at 19:48
APC89APC89
2,371720
asked Feb 1 at 19:48
APC89APC89
2,371720
asked Feb 1 at 19:48
APC89APC89
2,371720
2,371720
$begingroup$
the support of $Y_2 + X_2$ is $mathbb{R^{+}}$ and the support of $X_2$ is $mathbb{R^{+}}$, from this you can conclude that the support of $Y_2$ is $mathbb{R}$
$endgroup$
– rapidracim
Feb 1 at 20:02
$begingroup$
I thought so, but the problem is the marginal probability density function expression: when you make $y_{2}$ varies from $-infty$ to $+infty$, the integral diverges. So could you help me find out where is the mistake? By the way, if the one half disappears, we get an exponential random variable.
$endgroup$
– APC89
Feb 1 at 20:04
$begingroup$
btw have you tried finding the distribution of $-X_2$ and then use the convolution formula for the distribution of $Y_2$ ?
$endgroup$
– rapidracim
Feb 1 at 20:09
$begingroup$
We haven't studied the convolution formula yet.
$endgroup$
– APC89
Feb 1 at 20:10
1
$begingroup$
The solution of (c) should read $$f_{Y_2}(y_2) = int_{|y_2|}^infty frac{lambda^2}2exp(-lambda y_1)mathrm dy_1 = frac{lambda}2exp(-lambda |y_2|) $$ for every $y_2$ in $mathbb R$.
$endgroup$
– Did
Feb 1 at 20:13
|
show 1 more comment
$begingroup$
the support of $Y_2 + X_2$ is $mathbb{R^{+}}$ and the support of $X_2$ is $mathbb{R^{+}}$, from this you can conclude that the support of $Y_2$ is $mathbb{R}$
$endgroup$
– rapidracim
Feb 1 at 20:02
$begingroup$
I thought so, but the problem is the marginal probability density function expression: when you make $y_{2}$ varies from $-infty$ to $+infty$, the integral diverges. So could you help me find out where is the mistake? By the way, if the one half disappears, we get an exponential random variable.
$endgroup$
– APC89
Feb 1 at 20:04
$begingroup$
btw have you tried finding the distribution of $-X_2$ and then use the convolution formula for the distribution of $Y_2$ ?
$endgroup$
– rapidracim
Feb 1 at 20:09
$begingroup$
We haven't studied the convolution formula yet.
$endgroup$
– APC89
Feb 1 at 20:10
1
$begingroup$
The solution of (c) should read $$f_{Y_2}(y_2) = int_{|y_2|}^infty frac{lambda^2}2exp(-lambda y_1)mathrm dy_1 = frac{lambda}2exp(-lambda |y_2|) $$ for every $y_2$ in $mathbb R$.
$endgroup$
– Did
Feb 1 at 20:13
$begingroup$
the support of $Y_2 + X_2$ is $mathbb{R^{+}}$ and the support of $X_2$ is $mathbb{R^{+}}$, from this you can conclude that the support of $Y_2$ is $mathbb{R}$
$endgroup$
– rapidracim
Feb 1 at 20:02
$begingroup$
the support of $Y_2 + X_2$ is $mathbb{R^{+}}$ and the support of $X_2$ is $mathbb{R^{+}}$, from this you can conclude that the support of $Y_2$ is $mathbb{R}$
$endgroup$
– rapidracim
Feb 1 at 20:02
$begingroup$
I thought so, but the problem is the marginal probability density function expression: when you make $y_{2}$ varies from $-infty$ to $+infty$, the integral diverges. So could you help me find out where is the mistake? By the way, if the one half disappears, we get an exponential random variable.
$endgroup$
– APC89
Feb 1 at 20:04
$begingroup$
I thought so, but the problem is the marginal probability density function expression: when you make $y_{2}$ varies from $-infty$ to $+infty$, the integral diverges. So could you help me find out where is the mistake? By the way, if the one half disappears, we get an exponential random variable.
$endgroup$
– APC89
Feb 1 at 20:04
$begingroup$
btw have you tried finding the distribution of $-X_2$ and then use the convolution formula for the distribution of $Y_2$ ?
$endgroup$
– rapidracim
Feb 1 at 20:09
$begingroup$
btw have you tried finding the distribution of $-X_2$ and then use the convolution formula for the distribution of $Y_2$ ?
$endgroup$
– rapidracim
Feb 1 at 20:09
$begingroup$
We haven't studied the convolution formula yet.
$endgroup$
– APC89
Feb 1 at 20:10
$begingroup$
We haven't studied the convolution formula yet.
$endgroup$
– APC89
Feb 1 at 20:10
1
1
$begingroup$
The solution of (c) should read $$f_{Y_2}(y_2) = int_{|y_2|}^infty frac{lambda^2}2exp(-lambda y_1)mathrm dy_1 = frac{lambda}2exp(-lambda |y_2|) $$ for every $y_2$ in $mathbb R$.
$endgroup$
– Did
Feb 1 at 20:13
$begingroup$
The solution of (c) should read $$f_{Y_2}(y_2) = int_{|y_2|}^infty frac{lambda^2}2exp(-lambda y_1)mathrm dy_1 = frac{lambda}2exp(-lambda |y_2|) $$ for every $y_2$ in $mathbb R$.
$endgroup$
– Did
Feb 1 at 20:13
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
you must have $Y_1 ge Y_2$ AND $Y_1 ge - Y_2$
meaning you must have $Y_1 ge max(Y_2,-Y_2) = |Y_2| $
begin{align*}
f_{Y_{2}}(y_{2}) &= int_{-infty}^{+infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})mathrm{d}y_{1} \ &= int_{|y_2|}^{+infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})mathrm{d}y_{1} \
&= frac{lambda}{2}exp(-lambda |y_{2}|)
end{align*}
answered Feb 1 at 20:13
rapidracimrapidracim
1,7441419
$endgroup$
$begingroup$
Thank you for the contribution!
$endgroup$
– APC89
Feb 1 at 20:14
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%2f3096652%2ftransformation-of-random-variables-what-went-wrong%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
$begingroup$
you must have $Y_1 ge Y_2$ AND $Y_1 ge - Y_2$
meaning you must have $Y_1 ge max(Y_2,-Y_2) = |Y_2| $
begin{align*}
f_{Y_{2}}(y_{2}) &= int_{-infty}^{+infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})mathrm{d}y_{1} \ &= int_{|y_2|}^{+infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})mathrm{d}y_{1} \
&= frac{lambda}{2}exp(-lambda |y_{2}|)
end{align*}
answered Feb 1 at 20:13
rapidracimrapidracim
1,7441419
$endgroup$
$begingroup$
Thank you for the contribution!
$endgroup$
– APC89
Feb 1 at 20:14
add a comment |
$begingroup$
you must have $Y_1 ge Y_2$ AND $Y_1 ge - Y_2$
meaning you must have $Y_1 ge max(Y_2,-Y_2) = |Y_2| $
begin{align*}
f_{Y_{2}}(y_{2}) &= int_{-infty}^{+infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})mathrm{d}y_{1} \ &= int_{|y_2|}^{+infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})mathrm{d}y_{1} \
&= frac{lambda}{2}exp(-lambda |y_{2}|)
end{align*}
answered Feb 1 at 20:13
rapidracimrapidracim
1,7441419
$endgroup$
$begingroup$
Thank you for the contribution!
$endgroup$
– APC89
Feb 1 at 20:14
add a comment |
$begingroup$
you must have $Y_1 ge Y_2$ AND $Y_1 ge - Y_2$
meaning you must have $Y_1 ge max(Y_2,-Y_2) = |Y_2| $
begin{align*}
f_{Y_{2}}(y_{2}) &= int_{-infty}^{+infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})mathrm{d}y_{1} \ &= int_{|y_2|}^{+infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})mathrm{d}y_{1} \
&= frac{lambda}{2}exp(-lambda |y_{2}|)
end{align*}
answered Feb 1 at 20:13
rapidracimrapidracim
1,7441419
$endgroup$
you must have $Y_1 ge Y_2$ AND $Y_1 ge - Y_2$
meaning you must have $Y_1 ge max(Y_2,-Y_2) = |Y_2| $
begin{align*}
f_{Y_{2}}(y_{2}) &= int_{-infty}^{+infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})mathrm{d}y_{1} \ &= int_{|y_2|}^{+infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})mathrm{d}y_{1} \
&= frac{lambda}{2}exp(-lambda |y_{2}|)
end{align*}
answered Feb 1 at 20:13
rapidracimrapidracim
1,7441419
answered Feb 1 at 20:13
rapidracimrapidracim
1,7441419
answered Feb 1 at 20:13
rapidracimrapidracim
1,7441419
answered Feb 1 at 20:13
rapidracimrapidracim
1,7441419
1,7441419
$begingroup$
Thank you for the contribution!
$endgroup$
– APC89
Feb 1 at 20:14
add a comment |
$begingroup$
Thank you for the contribution!
$endgroup$
– APC89
Feb 1 at 20:14
$begingroup$
Thank you for the contribution!
$endgroup$
– APC89
Feb 1 at 20:14
$begingroup$
Thank you for the contribution!
$endgroup$
– APC89
Feb 1 at 20:14
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.
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%2f3096652%2ftransformation-of-random-variables-what-went-wrong%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
$begingroup$
the support of $Y_2 + X_2$ is $mathbb{R^{+}}$ and the support of $X_2$ is $mathbb{R^{+}}$, from this you can conclude that the support of $Y_2$ is $mathbb{R}$
$endgroup$
– rapidracim
Feb 1 at 20:02
$begingroup$
I thought so, but the problem is the marginal probability density function expression: when you make $y_{2}$ varies from $-infty$ to $+infty$, the integral diverges. So could you help me find out where is the mistake? By the way, if the one half disappears, we get an exponential random variable.
$endgroup$
– APC89
Feb 1 at 20:04
$begingroup$
btw have you tried finding the distribution of $-X_2$ and then use the convolution formula for the distribution of $Y_2$ ?
$endgroup$
– rapidracim
Feb 1 at 20:09
$begingroup$
We haven't studied the convolution formula yet.
$endgroup$
– APC89
Feb 1 at 20:10
1
$begingroup$
The solution of (c) should read $$f_{Y_2}(y_2) = int_{|y_2|}^infty frac{lambda^2}2exp(-lambda y_1)mathrm dy_1 = frac{lambda}2exp(-lambda |y_2|) $$ for every $y_2$ in $mathbb R$.
$endgroup$
– Did
Feb 1 at 20:13