Finding Density Function of Transformation
Let $Y$ be uniformly distributed over the interval $(−1, 3)$. Find the probability density function
of $U = Y^2$.
Attempted solution:
We know that
$$
Y=begin{cases}
1/4 && -1 leq y leq 3\
0 && text{elsewhere}
end{cases}
$$
We find the CDF of $U$ using $Y$
$$
P(U leq u) = P(Y^2 leq u) = P(-sqrt u leq Y leq sqrt u)
$$
$$
= int_{-sqrt u}^{sqrt u}frac{1}{4}dy = frac{sqrt u}{2}
$$
Therefore
$$
f(u) = frac{d}{du}frac{sqrt u}{2} = frac{1}{4sqrt u} qquad 0 leq u leq 9
$$
However, the answer is:
$$
f(u) = begin{cases} frac{1}{4sqrt u} && 0 leq u < 1\
frac{1}{8sqrt u} && 1 leq u leq 9
end{cases}
$$
Logically, since Y is between -1 and 3, this makes sense to me since there are two values of Y for which U will be between 0 and 1 for each value, and only one value of Y for which U is between 1 and 9.
Despite this, why didn't the computation pick up on this? Where did I go wrong?
probability-distributions
asked Jan 3 at 22:35
Bryden C
30418
add a comment |
Let $Y$ be uniformly distributed over the interval $(−1, 3)$. Find the probability density function
of $U = Y^2$.
Attempted solution:
We know that
$$
Y=begin{cases}
1/4 && -1 leq y leq 3\
0 && text{elsewhere}
end{cases}
$$
We find the CDF of $U$ using $Y$
$$
P(U leq u) = P(Y^2 leq u) = P(-sqrt u leq Y leq sqrt u)
$$
$$
= int_{-sqrt u}^{sqrt u}frac{1}{4}dy = frac{sqrt u}{2}
$$
Therefore
$$
f(u) = frac{d}{du}frac{sqrt u}{2} = frac{1}{4sqrt u} qquad 0 leq u leq 9
$$
However, the answer is:
$$
f(u) = begin{cases} frac{1}{4sqrt u} && 0 leq u < 1\
frac{1}{8sqrt u} && 1 leq u leq 9
end{cases}
$$
Logically, since Y is between -1 and 3, this makes sense to me since there are two values of Y for which U will be between 0 and 1 for each value, and only one value of Y for which U is between 1 and 9.
Despite this, why didn't the computation pick up on this? Where did I go wrong?
probability-distributions
asked Jan 3 at 22:35
Bryden C
30418
The statement of the problem, that you apparently copied, says "U= Y" but your calculation uses "$U= Y^2$. Which is it?
– user247327
Jan 3 at 22:42
My mistake. It should be $U = Y^2$. Sorry.
– Bryden C
Jan 3 at 22:44
The square function is not injective on $(-1,3)$ so taking the inversion as you do requires some care.
– LoveTooNap29
Jan 3 at 22:51
add a comment |
Let $Y$ be uniformly distributed over the interval $(−1, 3)$. Find the probability density function
of $U = Y^2$.
Attempted solution:
We know that
$$
Y=begin{cases}
1/4 && -1 leq y leq 3\
0 && text{elsewhere}
end{cases}
$$
We find the CDF of $U$ using $Y$
$$
P(U leq u) = P(Y^2 leq u) = P(-sqrt u leq Y leq sqrt u)
$$
$$
= int_{-sqrt u}^{sqrt u}frac{1}{4}dy = frac{sqrt u}{2}
$$
Therefore
$$
f(u) = frac{d}{du}frac{sqrt u}{2} = frac{1}{4sqrt u} qquad 0 leq u leq 9
$$
However, the answer is:
$$
f(u) = begin{cases} frac{1}{4sqrt u} && 0 leq u < 1\
frac{1}{8sqrt u} && 1 leq u leq 9
end{cases}
$$
Logically, since Y is between -1 and 3, this makes sense to me since there are two values of Y for which U will be between 0 and 1 for each value, and only one value of Y for which U is between 1 and 9.
Despite this, why didn't the computation pick up on this? Where did I go wrong?
probability-distributions
asked Jan 3 at 22:35
Bryden C
30418
Let $Y$ be uniformly distributed over the interval $(−1, 3)$. Find the probability density function
of $U = Y^2$.
Attempted solution:
We know that
$$
Y=begin{cases}
1/4 && -1 leq y leq 3\
0 && text{elsewhere}
end{cases}
$$
We find the CDF of $U$ using $Y$
$$
P(U leq u) = P(Y^2 leq u) = P(-sqrt u leq Y leq sqrt u)
$$
$$
= int_{-sqrt u}^{sqrt u}frac{1}{4}dy = frac{sqrt u}{2}
$$
Therefore
$$
f(u) = frac{d}{du}frac{sqrt u}{2} = frac{1}{4sqrt u} qquad 0 leq u leq 9
$$
However, the answer is:
$$
f(u) = begin{cases} frac{1}{4sqrt u} && 0 leq u < 1\
frac{1}{8sqrt u} && 1 leq u leq 9
end{cases}
$$
Logically, since Y is between -1 and 3, this makes sense to me since there are two values of Y for which U will be between 0 and 1 for each value, and only one value of Y for which U is between 1 and 9.
Despite this, why didn't the computation pick up on this? Where did I go wrong?
probability-distributions
probability-distributions
asked Jan 3 at 22:35
Bryden C
30418
asked Jan 3 at 22:35
Bryden C
30418
edited Jan 3 at 22:43
asked Jan 3 at 22:35
Bryden C
30418
asked Jan 3 at 22:35
Bryden C
30418
asked Jan 3 at 22:35
Bryden C
30418
30418
The statement of the problem, that you apparently copied, says "U= Y" but your calculation uses "$U= Y^2$. Which is it?
– user247327
Jan 3 at 22:42
My mistake. It should be $U = Y^2$. Sorry.
– Bryden C
Jan 3 at 22:44
The square function is not injective on $(-1,3)$ so taking the inversion as you do requires some care.
– LoveTooNap29
Jan 3 at 22:51
add a comment |
The statement of the problem, that you apparently copied, says "U= Y" but your calculation uses "$U= Y^2$. Which is it?
– user247327
Jan 3 at 22:42
My mistake. It should be $U = Y^2$. Sorry.
– Bryden C
Jan 3 at 22:44
The square function is not injective on $(-1,3)$ so taking the inversion as you do requires some care.
– LoveTooNap29
Jan 3 at 22:51
The statement of the problem, that you apparently copied, says "U= Y" but your calculation uses "$U= Y^2$. Which is it?
– user247327
Jan 3 at 22:42
The statement of the problem, that you apparently copied, says "U= Y" but your calculation uses "$U= Y^2$. Which is it?
– user247327
Jan 3 at 22:42
My mistake. It should be $U = Y^2$. Sorry.
– Bryden C
Jan 3 at 22:44
My mistake. It should be $U = Y^2$. Sorry.
– Bryden C
Jan 3 at 22:44
The square function is not injective on $(-1,3)$ so taking the inversion as you do requires some care.
– LoveTooNap29
Jan 3 at 22:51
The square function is not injective on $(-1,3)$ so taking the inversion as you do requires some care.
– LoveTooNap29
Jan 3 at 22:51
add a comment |
1 Answer
1
active
oldest
votes
If $1<u<9$, then
$$P(-sqrt{u}leq Y < sqrt{u}) = P(-1leq Y leq sqrt{u}) = int_{-1}^{sqrt{u}} 1/4 dy $$
This should solve your problem. Remember that Y has 0 density outside of the interval [-1,3].
answered Jan 3 at 22:52
Leander Tilsted Kristensen
112
New contributor
Leander Tilsted Kristensen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Wouldn't $Y=0$ satisfy your second probability range? If so, then $U = Y^2 = 0$ but $U not in (1,9)$.
– Bryden C
Jan 3 at 22:55
@BrydenC What you are saying does not make the answer wrong. The answer given by Leander Tilsted Kristensen is correct.
– Kavi Rama Murthy
2 days ago
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%2f3061099%2ffinding-density-function-of-transformation%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
If $1<u<9$, then
$$P(-sqrt{u}leq Y < sqrt{u}) = P(-1leq Y leq sqrt{u}) = int_{-1}^{sqrt{u}} 1/4 dy $$
This should solve your problem. Remember that Y has 0 density outside of the interval [-1,3].
answered Jan 3 at 22:52
Leander Tilsted Kristensen
112
New contributor
Leander Tilsted Kristensen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Wouldn't $Y=0$ satisfy your second probability range? If so, then $U = Y^2 = 0$ but $U not in (1,9)$.
– Bryden C
Jan 3 at 22:55
@BrydenC What you are saying does not make the answer wrong. The answer given by Leander Tilsted Kristensen is correct.
– Kavi Rama Murthy
2 days ago
add a comment |
If $1<u<9$, then
$$P(-sqrt{u}leq Y < sqrt{u}) = P(-1leq Y leq sqrt{u}) = int_{-1}^{sqrt{u}} 1/4 dy $$
This should solve your problem. Remember that Y has 0 density outside of the interval [-1,3].
answered Jan 3 at 22:52
Leander Tilsted Kristensen
112
New contributor
Leander Tilsted Kristensen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Wouldn't $Y=0$ satisfy your second probability range? If so, then $U = Y^2 = 0$ but $U not in (1,9)$.
– Bryden C
Jan 3 at 22:55
@BrydenC What you are saying does not make the answer wrong. The answer given by Leander Tilsted Kristensen is correct.
– Kavi Rama Murthy
2 days ago
add a comment |
If $1<u<9$, then
$$P(-sqrt{u}leq Y < sqrt{u}) = P(-1leq Y leq sqrt{u}) = int_{-1}^{sqrt{u}} 1/4 dy $$
This should solve your problem. Remember that Y has 0 density outside of the interval [-1,3].
answered Jan 3 at 22:52
Leander Tilsted Kristensen
112
New contributor
Leander Tilsted Kristensen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If $1<u<9$, then
$$P(-sqrt{u}leq Y < sqrt{u}) = P(-1leq Y leq sqrt{u}) = int_{-1}^{sqrt{u}} 1/4 dy $$
This should solve your problem. Remember that Y has 0 density outside of the interval [-1,3].
answered Jan 3 at 22:52
Leander Tilsted Kristensen
112
New contributor
Leander Tilsted Kristensen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 3 at 22:52
Leander Tilsted Kristensen
112
New contributor
Leander Tilsted Kristensen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 3 at 22:52
Leander Tilsted Kristensen
112
answered Jan 3 at 22:52
Leander Tilsted Kristensen
112
112
New contributor
Leander Tilsted Kristensen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Leander Tilsted Kristensen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Leander Tilsted Kristensen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Wouldn't $Y=0$ satisfy your second probability range? If so, then $U = Y^2 = 0$ but $U not in (1,9)$.
– Bryden C
Jan 3 at 22:55
@BrydenC What you are saying does not make the answer wrong. The answer given by Leander Tilsted Kristensen is correct.
– Kavi Rama Murthy
2 days ago
add a comment |
Wouldn't $Y=0$ satisfy your second probability range? If so, then $U = Y^2 = 0$ but $U not in (1,9)$.
– Bryden C
Jan 3 at 22:55
@BrydenC What you are saying does not make the answer wrong. The answer given by Leander Tilsted Kristensen is correct.
– Kavi Rama Murthy
2 days ago
Wouldn't $Y=0$ satisfy your second probability range? If so, then $U = Y^2 = 0$ but $U not in (1,9)$.
– Bryden C
Jan 3 at 22:55
Wouldn't $Y=0$ satisfy your second probability range? If so, then $U = Y^2 = 0$ but $U not in (1,9)$.
– Bryden C
Jan 3 at 22:55
@BrydenC What you are saying does not make the answer wrong. The answer given by Leander Tilsted Kristensen is correct.
– Kavi Rama Murthy
2 days ago
@BrydenC What you are saying does not make the answer wrong. The answer given by Leander Tilsted Kristensen is correct.
– Kavi Rama Murthy
2 days ago
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%2f3061099%2ffinding-density-function-of-transformation%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
The statement of the problem, that you apparently copied, says "U= Y" but your calculation uses "$U= Y^2$. Which is it?
– user247327
Jan 3 at 22:42
My mistake. It should be $U = Y^2$. Sorry.
– Bryden C
Jan 3 at 22:44
The square function is not injective on $(-1,3)$ so taking the inversion as you do requires some care.
– LoveTooNap29
Jan 3 at 22:51