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Let X have a uniform distribution on the interval $(0,1)$. Given that X = x, let Y have a uniform distribution on the interval $(0,x+1)$.

Find the joint pdf of X and Y. Sketch the region where $f(x,y) > 0$.

Find fY$(y)$, the marginal pdf of Y. Be sure to include the domain.

I'm not really sure where to start. Is fY$(y)$ just $\frac{1}{(x+1)-0}$ since that's the pdf of a uniform distribution? And I have no idea how to find the joint pdf of X and Y.

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The conditional density of $Y$ given $X=x$ is $f_{Y|X}(y|x)=\frac{1}{x+1}1_{0<y<x+1}$, hence the joint density of $X$ and $Y$ is $$ f(x,y)=f_{Y|X}(y|x)f_X(x)=\frac{1}{x+1}1_{0<x<1,0<y<x+1}$$

The marginal density can then be obtained by "integrating out" the $x$-variable: $$ f_Y(y)=\int f(x,y)\;dx$$

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  • $\begingroup$ Sorry, what do you mean by *f<sub>Y|X</sub>(y|x)*=1/x+1 1 <sub>0<y<x+1</sub>? $\endgroup$ – bhav1001 Apr 7 '16 at 3:32
  • $\begingroup$ This is the conditional density of $Y$ given $X=x$. See here for more info: en.wikipedia.org/wiki/Conditional_probability_distribution $\endgroup$ – carmichael561 Apr 7 '16 at 3:35
  • $\begingroup$ I think I get that, but why is there a 1 between 1/x+1 and the bounds? $\endgroup$ – bhav1001 Apr 7 '16 at 3:37
  • $\begingroup$ That is a shorthand notation for 1 if $0<y<x+1$, and 0 otherwise. $\endgroup$ – carmichael561 Apr 7 '16 at 3:41
  • $\begingroup$ I'm a bit confused on the bounds to use on the integral to calculate the marginal density of y. I believe $y$ should be the lower bound, but what is the upper bound? It can't be $x+1$, right? $\endgroup$ – bhav1001 Apr 7 '16 at 4:05

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