Find the conditional probability density function of $X$ given $Y=y$Find the marginal and conditional densities without explicitly having the joint density?Find the conditional and marginal densities?A joint density function with issues in the conditional density functionConditional Density and expectationHow to find the joint distribution and joint density functions of two random variables?Finding Probability Density Function and ProbabilityFind the probability density functions of $X$ and $Y$Computing conditional density of continuous rvRandom Variable $X$ and $Y$ has a joint probability density function. Find $f_ Y(x | y)$Find the conditional probability density function of $Y$ given $X=x$
Why can't the Brexit deadlock in the UK parliament be solved with a plurality vote?
Do I have to take mana from my deck or hand when tapping a dual land?
Would this string work as string?
In One Punch Man, is King actually weak?
How can I, as DM, avoid the Conga Line of Death occurring when implementing some form of flanking rule?
El Dorado Word Puzzle II: Videogame Edition
How to test the sharpness of a knife?
Does Doodling or Improvising on the Piano Have Any Benefits?
Why would five hundred and five be same as one?
Alignment of six matrices
Personal or impersonal in a technical resume
How do I Interface a PS/2 Keyboard without Modern Techniques?
Why the "ls" command is showing the permissions of files in a FAT32 partition?
Why didn't Voldemort know what Grindelwald looked like?
Do you waste sorcery points if you try to apply metamagic to a spell from a scroll but fail to cast it?
When and why was runway 07/25 at Kai Tak removed?
How do you justify more code being written by following clean code practices?
Check if object is null and return null
Should I assume I have passed probation?
Why does the Persian emissary display a string of crowned skulls?
Difference between shutdown options
Grepping string, but include all non-blank lines following each grep match
Is there a RAID 0 Equivalent for RAM?
Visualizing the difference curve in a 2D plot?
Find the conditional probability density function of $X$ given $Y=y$
Find the marginal and conditional densities without explicitly having the joint density?Find the conditional and marginal densities?A joint density function with issues in the conditional density functionConditional Density and expectationHow to find the joint distribution and joint density functions of two random variables?Finding Probability Density Function and ProbabilityFind the probability density functions of $X$ and $Y$Computing conditional density of continuous rvRandom Variable $X$ and $Y$ has a joint probability density function. Find $f_X (x | y)$Find the conditional probability density function of $Y$ given $X=x$
$begingroup$
Let $X$ and $Y$ be two random variables with joint density function:
$f_XY(x,y) = begincases 5x^2y&-1leq xleq1, 0<yleq|x| \0&textotherwise endcases$
Find $f_Y(x|y)$ the conditional probability density function of $X$ given $Y =
y$. Sketch the graph of $f_Y(x|.5)$
The graph part isn't the confusing issue as I feel once I get to it, I will be able to solve it. My issue is finding the proper $f_Y(x|y)$. I thought I would do $f_y(y)=int_-1^15x^2ydx$ and I get $10yover3$ then I plug that into the formula to get $5x^2yover10yover3$ $= 3x^2over2$
However, the answer is $3x^2over2(1-y^3)$
What am I doing wrong? I feel like it has to do with my limits of integration but I don't see how it isn't $-1$ to $1$.
probability density-function
edited Mar 14 at 6:14
Rócherz
2,9863821
asked Mar 7 '17 at 17:35
Heavenly96Heavenly96
279217
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be two random variables with joint density function:
$f_XY(x,y) = begincases 5x^2y&-1leq xleq1, 0<yleq|x| \0&textotherwise endcases$
Find $f_Y(x|y)$ the conditional probability density function of $X$ given $Y =
y$. Sketch the graph of $f_Y(x|.5)$
The graph part isn't the confusing issue as I feel once I get to it, I will be able to solve it. My issue is finding the proper $f_Y(x|y)$. I thought I would do $f_y(y)=int_-1^15x^2ydx$ and I get $10yover3$ then I plug that into the formula to get $5x^2yover10yover3$ $= 3x^2over2$
However, the answer is $3x^2over2(1-y^3)$
What am I doing wrong? I feel like it has to do with my limits of integration but I don't see how it isn't $-1$ to $1$.
probability density-function
edited Mar 14 at 6:14
Rócherz
2,9863821
asked Mar 7 '17 at 17:35
Heavenly96Heavenly96
279217
$endgroup$
3
$begingroup$
Your computation of $f_Y$ is wrong, actually, $$f_Y(y)=int_f_X,Y(x,y)dx=int_-1^-ycdots dx+int_y^1cdots dx$$
$endgroup$
– Did
Mar 7 '17 at 17:49
$begingroup$
@Did would you mind elaborating on why this is the case? For all the other problems, including some sample SOA problems, I did it the same way as I did in my attempt up above and got the correct answers. Is it due to this one involving $|x|$?
$endgroup$
– Heavenly96
Mar 7 '17 at 18:28
$begingroup$
The conditional density "answer" $3x^2over2(1-y^3)$ that you gave in the question is not right either unless you put in various constraints on the values that $x$ and $y$ take on.
$endgroup$
– Dilip Sarwate
Mar 7 '17 at 21:20
1
$begingroup$
This becomes more or less direct as soon as one writes down $f_X,Y$ as a true function defined on $mathbb R^2$, that is, $$f_X,Y(x,y)=5x^2ymathbf 1_<1$$ then the usual formula $$f_Y(y)=int_mathbb Rf_X,Y(x,y)dx$$ defines a true function $f_Y$ on $mathbb R$ and yields what I wrote in my first comment if $|y|<1$ and $0$ otherwise.
$endgroup$
– Did
Mar 7 '17 at 22:18
$begingroup$
*Typo: ...and yields what I wrote in my first comment if $0<y<1$ and $0$ otherwise.
$endgroup$
– Did
Mar 9 '17 at 9:58
add a comment |
$begingroup$
Let $X$ and $Y$ be two random variables with joint density function:
$f_XY(x,y) = begincases 5x^2y&-1leq xleq1, 0<yleq|x| \0&textotherwise endcases$
Find $f_Y(x|y)$ the conditional probability density function of $X$ given $Y =
y$. Sketch the graph of $f_Y(x|.5)$
The graph part isn't the confusing issue as I feel once I get to it, I will be able to solve it. My issue is finding the proper $f_Y(x|y)$. I thought I would do $f_y(y)=int_-1^15x^2ydx$ and I get $10yover3$ then I plug that into the formula to get $5x^2yover10yover3$ $= 3x^2over2$
However, the answer is $3x^2over2(1-y^3)$
What am I doing wrong? I feel like it has to do with my limits of integration but I don't see how it isn't $-1$ to $1$.
probability density-function
edited Mar 14 at 6:14
Rócherz
2,9863821
asked Mar 7 '17 at 17:35
Heavenly96Heavenly96
279217
$endgroup$
Let $X$ and $Y$ be two random variables with joint density function:
$f_XY(x,y) = begincases 5x^2y&-1leq xleq1, 0<yleq|x| \0&textotherwise endcases$
Find $f_Y(x|y)$ the conditional probability density function of $X$ given $Y =
y$. Sketch the graph of $f_Y(x|.5)$
The graph part isn't the confusing issue as I feel once I get to it, I will be able to solve it. My issue is finding the proper $f_Y(x|y)$. I thought I would do $f_y(y)=int_-1^15x^2ydx$ and I get $10yover3$ then I plug that into the formula to get $5x^2yover10yover3$ $= 3x^2over2$
However, the answer is $3x^2over2(1-y^3)$
What am I doing wrong? I feel like it has to do with my limits of integration but I don't see how it isn't $-1$ to $1$.
probability density-function
probability density-function
edited Mar 14 at 6:14
Rócherz
2,9863821
asked Mar 7 '17 at 17:35
Heavenly96Heavenly96
279217
edited Mar 14 at 6:14
Rócherz
2,9863821
asked Mar 7 '17 at 17:35
Heavenly96Heavenly96
279217
edited Mar 14 at 6:14
Rócherz
2,9863821
edited Mar 14 at 6:14
Rócherz
2,9863821
edited Mar 14 at 6:14
Rócherz
2,9863821
2,9863821
asked Mar 7 '17 at 17:35
Heavenly96Heavenly96
279217
asked Mar 7 '17 at 17:35
Heavenly96Heavenly96
279217
asked Mar 7 '17 at 17:35
Heavenly96Heavenly96
279217
279217
3
$begingroup$
Your computation of $f_Y$ is wrong, actually, $$f_Y(y)=int_f_X,Y(x,y)dx=int_-1^-ycdots dx+int_y^1cdots dx$$
$endgroup$
– Did
Mar 7 '17 at 17:49
$begingroup$
@Did would you mind elaborating on why this is the case? For all the other problems, including some sample SOA problems, I did it the same way as I did in my attempt up above and got the correct answers. Is it due to this one involving $|x|$?
$endgroup$
– Heavenly96
Mar 7 '17 at 18:28
$begingroup$
The conditional density "answer" $3x^2over2(1-y^3)$ that you gave in the question is not right either unless you put in various constraints on the values that $x$ and $y$ take on.
$endgroup$
– Dilip Sarwate
Mar 7 '17 at 21:20
1
$begingroup$
This becomes more or less direct as soon as one writes down $f_X,Y$ as a true function defined on $mathbb R^2$, that is, $$f_X,Y(x,y)=5x^2ymathbf 1_<1$$ then the usual formula $$f_Y(y)=int_mathbb Rf_X,Y(x,y)dx$$ defines a true function $f_Y$ on $mathbb R$ and yields what I wrote in my first comment if $|y|<1$ and $0$ otherwise.
$endgroup$
– Did
Mar 7 '17 at 22:18
$begingroup$
*Typo: ...and yields what I wrote in my first comment if $0<y<1$ and $0$ otherwise.
$endgroup$
– Did
Mar 9 '17 at 9:58
add a comment |
3
$begingroup$
Your computation of $f_Y$ is wrong, actually, $$f_Y(y)=int_f_X,Y(x,y)dx=int_-1^-ycdots dx+int_y^1cdots dx$$
$endgroup$
– Did
Mar 7 '17 at 17:49
$begingroup$
@Did would you mind elaborating on why this is the case? For all the other problems, including some sample SOA problems, I did it the same way as I did in my attempt up above and got the correct answers. Is it due to this one involving $|x|$?
$endgroup$
– Heavenly96
Mar 7 '17 at 18:28
$begingroup$
The conditional density "answer" $3x^2over2(1-y^3)$ that you gave in the question is not right either unless you put in various constraints on the values that $x$ and $y$ take on.
$endgroup$
– Dilip Sarwate
Mar 7 '17 at 21:20
1
$begingroup$
This becomes more or less direct as soon as one writes down $f_X,Y$ as a true function defined on $mathbb R^2$, that is, $$f_X,Y(x,y)=5x^2ymathbf 1_<1$$ then the usual formula $$f_Y(y)=int_mathbb Rf_X,Y(x,y)dx$$ defines a true function $f_Y$ on $mathbb R$ and yields what I wrote in my first comment if $|y|<1$ and $0$ otherwise.
$endgroup$
– Did
Mar 7 '17 at 22:18
$begingroup$
*Typo: ...and yields what I wrote in my first comment if $0<y<1$ and $0$ otherwise.
$endgroup$
– Did
Mar 9 '17 at 9:58
3
3
$begingroup$
Your computation of $f_Y$ is wrong, actually, $$f_Y(y)=int_f_X,Y(x,y)dx=int_-1^-ycdots dx+int_y^1cdots dx$$
$endgroup$
– Did
Mar 7 '17 at 17:49
$begingroup$
Your computation of $f_Y$ is wrong, actually, $$f_Y(y)=int_f_X,Y(x,y)dx=int_-1^-ycdots dx+int_y^1cdots dx$$
$endgroup$
– Did
Mar 7 '17 at 17:49
$begingroup$
@Did would you mind elaborating on why this is the case? For all the other problems, including some sample SOA problems, I did it the same way as I did in my attempt up above and got the correct answers. Is it due to this one involving $|x|$?
$endgroup$
– Heavenly96
Mar 7 '17 at 18:28
$begingroup$
@Did would you mind elaborating on why this is the case? For all the other problems, including some sample SOA problems, I did it the same way as I did in my attempt up above and got the correct answers. Is it due to this one involving $|x|$?
$endgroup$
– Heavenly96
Mar 7 '17 at 18:28
$begingroup$
The conditional density "answer" $3x^2over2(1-y^3)$ that you gave in the question is not right either unless you put in various constraints on the values that $x$ and $y$ take on.
$endgroup$
– Dilip Sarwate
Mar 7 '17 at 21:20
$begingroup$
The conditional density "answer" $3x^2over2(1-y^3)$ that you gave in the question is not right either unless you put in various constraints on the values that $x$ and $y$ take on.
$endgroup$
– Dilip Sarwate
Mar 7 '17 at 21:20
1
1
$begingroup$
This becomes more or less direct as soon as one writes down $f_X,Y$ as a true function defined on $mathbb R^2$, that is, $$f_X,Y(x,y)=5x^2ymathbf 1_<1$$ then the usual formula $$f_Y(y)=int_mathbb Rf_X,Y(x,y)dx$$ defines a true function $f_Y$ on $mathbb R$ and yields what I wrote in my first comment if $|y|<1$ and $0$ otherwise.
$endgroup$
– Did
Mar 7 '17 at 22:18
$begingroup$
This becomes more or less direct as soon as one writes down $f_X,Y$ as a true function defined on $mathbb R^2$, that is, $$f_X,Y(x,y)=5x^2ymathbf 1_<1$$ then the usual formula $$f_Y(y)=int_mathbb Rf_X,Y(x,y)dx$$ defines a true function $f_Y$ on $mathbb R$ and yields what I wrote in my first comment if $|y|<1$ and $0$ otherwise.
$endgroup$
– Did
Mar 7 '17 at 22:18
$begingroup$
*Typo: ...and yields what I wrote in my first comment if $0<y<1$ and $0$ otherwise.
$endgroup$
– Did
Mar 9 '17 at 9:58
$begingroup$
*Typo: ...and yields what I wrote in my first comment if $0<y<1$ and $0$ otherwise.
$endgroup$
– Did
Mar 9 '17 at 9:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First, sketch the support of $(X,Y)$, which is $$(-1 le X le 1) cap (0 < Y le |X|).$$ This describes a "bow-tie" shaped region in the Cartesian coordinate plane, consisting of two right triangles with vertices $$(0,0), (1,0), (1,1), quad (0,0), (-1,0), (-1,1).$$ The symmetry of this region is clear, and the line of symmetry is the $y$-axis.
The plot below visualizes the joint density in three-dimensional space:
For a fixed $Y = y$, the permissible values of $X$ comprise the union of two intervals; that is to say, $$(X mid Y = y) in [-1, -y] cup [y, 1].$$ So, the interval of integration for the marginal density of $Y$ is not $[-1,1]$ but the interval above: $$f_Y(y) = int_x=-1^-y 5x^2 y , dx + int_x=y^1 5x^2 y , dx.$$
The above animation plots $X$ for specific values of $Y = y$ for $y in (0,1]$. The conditional density $f_Xmid Y(x mid y)$ is proportional to this plot in such a way that the area under the curve is made equal to $1$. In other words, the function $f_X mid Y$ is simply a scaled version of this animation such that the area under the curve remains $1$ regardless of the choice of $y$:
Compare this with the previous animation.
Another way to reason about this is to observe that the support requires that $|X| ge Y$: hence, on the interval $X in [-y,y]$, $f_X,Y(x,y) = 0$, thus writing $$f_Y(y) = int_x=-1^1 f_X,Y(x,y) , dx = int_x=-1^1 5x^2 y , dx$$ fails to reflect that the joint density is not $5x^2 y$ when $-y le x le y$. We can correct for this by the suitable use of indicator functions; for example, we could have instead written the joint density as $$f_X,Y(x,y) = 5x^2 y ;mathbb 1(|x| ge y), quad (x,y) in [-1,1] times [0,1].$$ This reminds us that integration in the rectangular region $[-1,1] times [0,1]$ comes with an extra condition for the integrand to be positive: $$f_Y(y) = int_x=-1^1 5x^2 y ; mathbb 1(|x| ge y) , dx = int_x=-1^-y 5x^2 y , dx + int_x=y^1 5x^2 y , dx.$$ We could go overboard and write the whole density as $$f_X,Y(x,y) = 5x^2 y ; mathbb 1 (0 < y le |x| le 1), quad (x,y) in mathbb R^2$$ but this isn't really necessary as we already intuitively grasp that the convex boundary of the support is rectangular, leading to a natural choice of the lower and upper limits of integration except when dealing with the nonconvex part.
answered Mar 7 '17 at 21:20
heropupheropup
64.6k764103
$endgroup$
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%2f2176285%2ffind-the-conditional-probability-density-function-of-x-given-y-y%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
$begingroup$
First, sketch the support of $(X,Y)$, which is $$(-1 le X le 1) cap (0 < Y le |X|).$$ This describes a "bow-tie" shaped region in the Cartesian coordinate plane, consisting of two right triangles with vertices $$(0,0), (1,0), (1,1), quad (0,0), (-1,0), (-1,1).$$ The symmetry of this region is clear, and the line of symmetry is the $y$-axis.
The plot below visualizes the joint density in three-dimensional space:
For a fixed $Y = y$, the permissible values of $X$ comprise the union of two intervals; that is to say, $$(X mid Y = y) in [-1, -y] cup [y, 1].$$ So, the interval of integration for the marginal density of $Y$ is not $[-1,1]$ but the interval above: $$f_Y(y) = int_x=-1^-y 5x^2 y , dx + int_x=y^1 5x^2 y , dx.$$
The above animation plots $X$ for specific values of $Y = y$ for $y in (0,1]$. The conditional density $f_Xmid Y(x mid y)$ is proportional to this plot in such a way that the area under the curve is made equal to $1$. In other words, the function $f_X mid Y$ is simply a scaled version of this animation such that the area under the curve remains $1$ regardless of the choice of $y$:
Compare this with the previous animation.
Another way to reason about this is to observe that the support requires that $|X| ge Y$: hence, on the interval $X in [-y,y]$, $f_X,Y(x,y) = 0$, thus writing $$f_Y(y) = int_x=-1^1 f_X,Y(x,y) , dx = int_x=-1^1 5x^2 y , dx$$ fails to reflect that the joint density is not $5x^2 y$ when $-y le x le y$. We can correct for this by the suitable use of indicator functions; for example, we could have instead written the joint density as $$f_X,Y(x,y) = 5x^2 y ;mathbb 1(|x| ge y), quad (x,y) in [-1,1] times [0,1].$$ This reminds us that integration in the rectangular region $[-1,1] times [0,1]$ comes with an extra condition for the integrand to be positive: $$f_Y(y) = int_x=-1^1 5x^2 y ; mathbb 1(|x| ge y) , dx = int_x=-1^-y 5x^2 y , dx + int_x=y^1 5x^2 y , dx.$$ We could go overboard and write the whole density as $$f_X,Y(x,y) = 5x^2 y ; mathbb 1 (0 < y le |x| le 1), quad (x,y) in mathbb R^2$$ but this isn't really necessary as we already intuitively grasp that the convex boundary of the support is rectangular, leading to a natural choice of the lower and upper limits of integration except when dealing with the nonconvex part.
answered Mar 7 '17 at 21:20
heropupheropup
64.6k764103
$endgroup$
add a comment |
$begingroup$
First, sketch the support of $(X,Y)$, which is $$(-1 le X le 1) cap (0 < Y le |X|).$$ This describes a "bow-tie" shaped region in the Cartesian coordinate plane, consisting of two right triangles with vertices $$(0,0), (1,0), (1,1), quad (0,0), (-1,0), (-1,1).$$ The symmetry of this region is clear, and the line of symmetry is the $y$-axis.
The plot below visualizes the joint density in three-dimensional space:
For a fixed $Y = y$, the permissible values of $X$ comprise the union of two intervals; that is to say, $$(X mid Y = y) in [-1, -y] cup [y, 1].$$ So, the interval of integration for the marginal density of $Y$ is not $[-1,1]$ but the interval above: $$f_Y(y) = int_x=-1^-y 5x^2 y , dx + int_x=y^1 5x^2 y , dx.$$
The above animation plots $X$ for specific values of $Y = y$ for $y in (0,1]$. The conditional density $f_Xmid Y(x mid y)$ is proportional to this plot in such a way that the area under the curve is made equal to $1$. In other words, the function $f_X mid Y$ is simply a scaled version of this animation such that the area under the curve remains $1$ regardless of the choice of $y$:
Compare this with the previous animation.
Another way to reason about this is to observe that the support requires that $|X| ge Y$: hence, on the interval $X in [-y,y]$, $f_X,Y(x,y) = 0$, thus writing $$f_Y(y) = int_x=-1^1 f_X,Y(x,y) , dx = int_x=-1^1 5x^2 y , dx$$ fails to reflect that the joint density is not $5x^2 y$ when $-y le x le y$. We can correct for this by the suitable use of indicator functions; for example, we could have instead written the joint density as $$f_X,Y(x,y) = 5x^2 y ;mathbb 1(|x| ge y), quad (x,y) in [-1,1] times [0,1].$$ This reminds us that integration in the rectangular region $[-1,1] times [0,1]$ comes with an extra condition for the integrand to be positive: $$f_Y(y) = int_x=-1^1 5x^2 y ; mathbb 1(|x| ge y) , dx = int_x=-1^-y 5x^2 y , dx + int_x=y^1 5x^2 y , dx.$$ We could go overboard and write the whole density as $$f_X,Y(x,y) = 5x^2 y ; mathbb 1 (0 < y le |x| le 1), quad (x,y) in mathbb R^2$$ but this isn't really necessary as we already intuitively grasp that the convex boundary of the support is rectangular, leading to a natural choice of the lower and upper limits of integration except when dealing with the nonconvex part.
answered Mar 7 '17 at 21:20
heropupheropup
64.6k764103
$endgroup$
add a comment |
$begingroup$
First, sketch the support of $(X,Y)$, which is $$(-1 le X le 1) cap (0 < Y le |X|).$$ This describes a "bow-tie" shaped region in the Cartesian coordinate plane, consisting of two right triangles with vertices $$(0,0), (1,0), (1,1), quad (0,0), (-1,0), (-1,1).$$ The symmetry of this region is clear, and the line of symmetry is the $y$-axis.
The plot below visualizes the joint density in three-dimensional space:
For a fixed $Y = y$, the permissible values of $X$ comprise the union of two intervals; that is to say, $$(X mid Y = y) in [-1, -y] cup [y, 1].$$ So, the interval of integration for the marginal density of $Y$ is not $[-1,1]$ but the interval above: $$f_Y(y) = int_x=-1^-y 5x^2 y , dx + int_x=y^1 5x^2 y , dx.$$
The above animation plots $X$ for specific values of $Y = y$ for $y in (0,1]$. The conditional density $f_Xmid Y(x mid y)$ is proportional to this plot in such a way that the area under the curve is made equal to $1$. In other words, the function $f_X mid Y$ is simply a scaled version of this animation such that the area under the curve remains $1$ regardless of the choice of $y$:
Compare this with the previous animation.
Another way to reason about this is to observe that the support requires that $|X| ge Y$: hence, on the interval $X in [-y,y]$, $f_X,Y(x,y) = 0$, thus writing $$f_Y(y) = int_x=-1^1 f_X,Y(x,y) , dx = int_x=-1^1 5x^2 y , dx$$ fails to reflect that the joint density is not $5x^2 y$ when $-y le x le y$. We can correct for this by the suitable use of indicator functions; for example, we could have instead written the joint density as $$f_X,Y(x,y) = 5x^2 y ;mathbb 1(|x| ge y), quad (x,y) in [-1,1] times [0,1].$$ This reminds us that integration in the rectangular region $[-1,1] times [0,1]$ comes with an extra condition for the integrand to be positive: $$f_Y(y) = int_x=-1^1 5x^2 y ; mathbb 1(|x| ge y) , dx = int_x=-1^-y 5x^2 y , dx + int_x=y^1 5x^2 y , dx.$$ We could go overboard and write the whole density as $$f_X,Y(x,y) = 5x^2 y ; mathbb 1 (0 < y le |x| le 1), quad (x,y) in mathbb R^2$$ but this isn't really necessary as we already intuitively grasp that the convex boundary of the support is rectangular, leading to a natural choice of the lower and upper limits of integration except when dealing with the nonconvex part.
answered Mar 7 '17 at 21:20
heropupheropup
64.6k764103
$endgroup$
First, sketch the support of $(X,Y)$, which is $$(-1 le X le 1) cap (0 < Y le |X|).$$ This describes a "bow-tie" shaped region in the Cartesian coordinate plane, consisting of two right triangles with vertices $$(0,0), (1,0), (1,1), quad (0,0), (-1,0), (-1,1).$$ The symmetry of this region is clear, and the line of symmetry is the $y$-axis.
The plot below visualizes the joint density in three-dimensional space:
For a fixed $Y = y$, the permissible values of $X$ comprise the union of two intervals; that is to say, $$(X mid Y = y) in [-1, -y] cup [y, 1].$$ So, the interval of integration for the marginal density of $Y$ is not $[-1,1]$ but the interval above: $$f_Y(y) = int_x=-1^-y 5x^2 y , dx + int_x=y^1 5x^2 y , dx.$$
The above animation plots $X$ for specific values of $Y = y$ for $y in (0,1]$. The conditional density $f_Xmid Y(x mid y)$ is proportional to this plot in such a way that the area under the curve is made equal to $1$. In other words, the function $f_X mid Y$ is simply a scaled version of this animation such that the area under the curve remains $1$ regardless of the choice of $y$:
Compare this with the previous animation.
Another way to reason about this is to observe that the support requires that $|X| ge Y$: hence, on the interval $X in [-y,y]$, $f_X,Y(x,y) = 0$, thus writing $$f_Y(y) = int_x=-1^1 f_X,Y(x,y) , dx = int_x=-1^1 5x^2 y , dx$$ fails to reflect that the joint density is not $5x^2 y$ when $-y le x le y$. We can correct for this by the suitable use of indicator functions; for example, we could have instead written the joint density as $$f_X,Y(x,y) = 5x^2 y ;mathbb 1(|x| ge y), quad (x,y) in [-1,1] times [0,1].$$ This reminds us that integration in the rectangular region $[-1,1] times [0,1]$ comes with an extra condition for the integrand to be positive: $$f_Y(y) = int_x=-1^1 5x^2 y ; mathbb 1(|x| ge y) , dx = int_x=-1^-y 5x^2 y , dx + int_x=y^1 5x^2 y , dx.$$ We could go overboard and write the whole density as $$f_X,Y(x,y) = 5x^2 y ; mathbb 1 (0 < y le |x| le 1), quad (x,y) in mathbb R^2$$ but this isn't really necessary as we already intuitively grasp that the convex boundary of the support is rectangular, leading to a natural choice of the lower and upper limits of integration except when dealing with the nonconvex part.
answered Mar 7 '17 at 21:20
heropupheropup
64.6k764103
edited Mar 7 '17 at 21:35
answered Mar 7 '17 at 21:20
heropupheropup
64.6k764103
answered Mar 7 '17 at 21:20
heropupheropup
64.6k764103
answered Mar 7 '17 at 21:20
heropupheropup
64.6k764103
64.6k764103
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.
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%2f2176285%2ffind-the-conditional-probability-density-function-of-x-given-y-y%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
3
$begingroup$
Your computation of $f_Y$ is wrong, actually, $$f_Y(y)=int_f_X,Y(x,y)dx=int_-1^-ycdots dx+int_y^1cdots dx$$
$endgroup$
– Did
Mar 7 '17 at 17:49
$begingroup$
@Did would you mind elaborating on why this is the case? For all the other problems, including some sample SOA problems, I did it the same way as I did in my attempt up above and got the correct answers. Is it due to this one involving $|x|$?
$endgroup$
– Heavenly96
Mar 7 '17 at 18:28
$begingroup$
The conditional density "answer" $3x^2over2(1-y^3)$ that you gave in the question is not right either unless you put in various constraints on the values that $x$ and $y$ take on.
$endgroup$
– Dilip Sarwate
Mar 7 '17 at 21:20
1
$begingroup$
This becomes more or less direct as soon as one writes down $f_X,Y$ as a true function defined on $mathbb R^2$, that is, $$f_X,Y(x,y)=5x^2ymathbf 1_<1$$ then the usual formula $$f_Y(y)=int_mathbb Rf_X,Y(x,y)dx$$ defines a true function $f_Y$ on $mathbb R$ and yields what I wrote in my first comment if $|y|<1$ and $0$ otherwise.
$endgroup$
– Did
Mar 7 '17 at 22:18
$begingroup$
*Typo: ...and yields what I wrote in my first comment if $0<y<1$ and $0$ otherwise.
$endgroup$
– Did
Mar 9 '17 at 9:58