I am faced with the following question, which I think is quite simple, but I can't put together for some reason.
Given that $f(x,y)=(6/5)(x+y^2)$ for $0<x,y<1$, ($f(x,y)=0$ everywhere else), I am asked to find both marginal densities and use them to find the probabilities that $X>0.8$ and that $Y<0.5$ Can someone here help me to understand the definition of marginal density, and show me how best to take a crack at this problem?
Thanks.
The marginal density function of $X$ is the plain old density function of $X$, nothing more, nothing less.
As to how to find the (marginal) density of $X$, we take a bit of a detour. Let us find the cumulative distribution function $F_X(x)$ of $X$. This is $\Pr(X\le x)$, and $$\Pr(X\le x)=\int_{x=-\infty}^x \left(\int_{-\infty}^\infty f_{X,Y}(x,y)\,dy\right)\,dx$$ where $f_{X,Y}(x,y)$ is the joint density of $X$ and $Y$. To find the density function of $X$, differentiate the cdf $F_X(x)$. By the Fundamental Theorem of Calculus, we have $$f_X(x)=\int_{-\infty}^\infty f_{X,Y}(x,y)\,dy.$$
Much more briefly, to find the (marginal) density of $X$, "integrate out" $y$.
To find the (marginal) density of $Y$, integrate out $x$.
Now for the calculations. We confine attention to where the action is, the unit square. We have $$f_X(x)=\int_0^1 \frac{6}{5}(x^2+y)\,dy$$ and $$f_Y(y)=\int_0^1 \frac{6}{5}(x^2+y)\,dx$$
Both integrations are straightforward. For accuracy, one should note that the densities are as obtained by the integrations on $(0,1)$, and $0$ elsewhere.