probabilityrandom variablestriangles
Let the random variables $X$ and $Y$ have a joint PDF which is uniform
over the triangle with vertices at $(0, 0), (0, 1 )$ and $(1, 0)$.
- Find the joint PDF of $X$ and $Y$.
So apparently the answer is $2$. And it's related with the area of this triangle that easily we can see is $1/2$. How do we get that answer? I guess I don't understand the meaning of joint PDF. Could someone explain that to me too?
For any $x, y$ in $[0, 1]$, \begin{align} \Pr(X \le x, Y \le y) & = \Pr(\sqrt U \le x, UV \le y) \\ & = \Pr\left(\sqrt U \le x, U \le \frac yV\right) \\ & = \int_0^1 \Pr\left(U \le x^2, U \le \frac yV \mid V = v\right) dv \\ & = \int_0^1 \min\left\{x^2, \frac yv\right\} dv \\ & = \begin{cases} x^2 & ; x^2 \le y \\ \int_0^{y/x^2} x^2 dv + \int_{y/x^2}^1 \frac yv dv & ; x^2 \ge y \\ \end{cases}\\ & = \begin{cases} x^2 & ; x^2 \le y \\ y - y \log y + 2y\log x & ; x^2 \ge y \\ \end{cases} \end{align} To obtain the probability density function, you take the derivative with respect to $x$ and $y$: $$ f_{X,Y}(x,y) = \begin{cases} 0 & ; x^2 \le y \\ 2/x & ; x^2 \ge y \\ \end{cases} $$
Here is a sketch of the support of the pdf and the cdf, and a geometrical computation of $F_{X,Y}(x_0,y_0)=\frac12y_0-\frac14x_0^2$ (check it, it is a good exercice).
It can help to understand why, for example concerning the marginal in $X$, we integrate with respect to $y$ (this is general) with the integration bounds $x_0$ and $2$ (this is particular):
$$f_X(x_0)=\int_{y=x_0}^{y=2}f(x,y)dy=\int_{y=x_0}^{y=2}\dfrac12dy=\dfrac12(2-x_0)$$
and a similar computation for $f_Y(y_0)=\int_{x=0}^{x=y_0}f(x,y)dy=\cdots.$
Best Answer
The random variable $(X,Y)$ is uniformly distributed over the region, lets call it $R$, i.e. the pdf $f_{X,Y}(x,y) = k$ for some constant on the region. Now lets integrate over the region
$$\int\int_R f_{X,Y}(x,y)dxdy = \int\int_R k \:dxdy$$
$$\int\int_R f_{X,Y}(x,y)dxdy = k * area(R)$$
$$\int\int_R f_{X,Y}(x,y)dxdy = k * \frac{1}{2}$$
We also know that
$$\int\int_R f_{X,Y}(x,y)dxdy = 1$$
so
$$k * \frac{1}{2} = 1 \Rightarrow k = 2 $$
And we end up with this pdf:
$$f_{X,Y}(x,y) = \begin{cases} 2, & (x,y) \in R \\ 0, & \text{otherwise} \end{cases} $$