integrationmarginal-distributionprobability distributions
I am having trouble finding the marginal distribution of $Y$ in the example below:
Let the joint pdf of $(X,Y)$ be $f(x,y) = 1$, for $0<x<1$ and $x<y<x+1$.
I tried something like $$\int_0^1 1 \, dx $$ but I think I got the integration bounds wrong. Any help would be greatly appreciated. Thanks very much!
The indicator is clear...what is wrong is that you have $f(u,v)$ and not $f(x,y)$
The support is a parallelogram (do a drawing to realize that) with area $1$ thus your density is uniform over this parallelogram and $C=1$, the reciprocal of the support area. No integrals are needed.
To get the two marginals you have to integrate
That is $U\sim U(0;1)$
For $V$ you have to observe that,
$$f_V(v)=\int_0^v du=v$$
$$f_V(v)=\int_{v-1}^1 du=2-v$$
concluding, $f_V$ is a triangular density
This is a compact way to write it
$$f_V(v)=[1-|1-v|]\mathbb{1}_{(0;2)}(v)$$
Best Answer
This may be a really cumbersome way (or perhaps not a correct way of doing it).
I hope you noted the region is a parallelogram and the probability is uniformly distributed on this parallelogram. Then if you were to find the CDF, you can essentially finding the area.
What I found was $F_Y(y)=\frac{y^2}{2}$ if $0\leq y\leq 1$ whereas $F_Y(y)=1-\frac{(2-y)^2}{2}$ for $2\geq y \geq 1$. Differentiate if you need to get pdf.