density functionmarginal-distributionprobability distributions
The Question
So I have tried the formula
$f_{X,Y}(x,y)=f_{Y\mid X}(y\mid x)\cdot f_{X}(x)$ but I am not able to figure out the bounds and I am not sure what to do after the bounds are found
Please do let me know if you guys have any idea on how to solve this question
since, for $n \in \mathbb{N}$ $$ p_n = e^{-\lambda}\frac{\lambda^n}{n!} $$ form the probability generating function: $$ f(x) = \sum_{n=0}^{\infty}p_nx^n= \sum_{n=0}^{\infty}e^{-\lambda}\frac{\lambda^n}{n!}x^n = e^{\lambda(x-1)} $$ now we have for the first and second moments: $$ \mu_1 = \sum_{n=0}^{\infty}np_n=f'(1) = \lambda \\ \mu_2 = \sum_{n=0}^{\infty}n^2p_n=\frac{d}{dx}(xf'(x))|_{x=1}=f'(1)+f''(1) =\lambda+\lambda^2 $$ so finally, for the standard deviation $$ \sigma^2=\mu_2-\mu_1^2 = \lambda $$
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
$$f_{XY}(x,y)=f_X(x)\cdot f_{Y|X}(y|x)=2x^2\mathbb{1}_{(0;1)}(x)\cdot\mathbb{1}_{(0;1/x)}(y)$$
now to calculate $f_Y(y)$ you have to integrate in X, that is
$$f_Y(y)=\int_0^1 2x^2dx=2/3$$
when $y \in[0;1]$
and
$$f_Y(y)=\int_0^{1/y} 2x^2 dx=\frac{2}{3y^3}$$
when $y>1$
To understand integral bounds it is useful to do a drawing of the integration area, first