Suppose that $(X_1, X_2)$ has joint pdf $f(x_1,x_2)=24x_1x_2$ with $1>x_1, x_2>0$ and $x_1+x_2<1$.
(1) Find $P(X_1<X_2)$.
(2) Find the marginal of $X_1$ and $X_2$.
I can get the probability by the area of the graph, which is the fourth of the square (=1/4). I try to calculate this integral by formula:
$$
\int\int_C (24x_1x_2)dx_1dx_2
$$
where $C=\{1>x_1, x_2>0, x_1+x_2<1, x_1<x_2\}$.
But how to calculate the above double integral and marginal integral?
For the first double integral,
$$
\int\int_C (24x_1x_2)dx_2dx_2=\int_0^{1/2}dx_1 \int_{x_2=1-x_1}^{x_1} (24x_1x_2)dx_2=12\int_0^{1/2}(2x_1^2-x_1)dx_1
$$
But I got $1/2$ but not $1/4$…
For (b), the marginal of $X_2$ is given by
$$
f_{X_2}(x)=\int_0^1 f(x_1, x_2)dx_1=
$$
I am not sure why the above integral is equal to $$\int_0^{1-x_2} (24x_1x_2)dx_1$$
Best Answer
(1) $$\iint_C (24x_1x_2)\,dx_{\color\red1}dx_2=\int_0^{1/2}\,dx_1\int_{x_2=1-x_1}^{x_1} (24x_1x_2)\,dx_2=12\int_0^{1/2}(x_1-2x_1^2)\,dx_1=\frac12,$$ which is not surprising since $P(X_1<X_2)=P(X_2<X_1).$
(2) The "marginal" density of $X_2$ is $$f_{X_2}(x_2)=\int_{0<x_1,x_2\atop x_1+x_2<1}24x_1x_2\, dx_1=24x_2{\bf1}_{(0,1)}(x_2)\int_0^{1-x_2}x_1\,dx_1=12x_2(1-x_2)^2{\bf1}_{(0,1)}(x_2),$$ where ${\bf1}_{(0,1)}$ is the indicator function of the interval $(0,1).$ You can check that $\int_0^112x_2(1-x_2)^2\,dx_2=1.$