Let $$f:\mathbb{R}^2 \to \mathbb{R}, (x,y) \mapsto \ f(x,y)=\left\{\begin{array}{ll} N\cdot \mathrm{e}^{-\alpha x-\beta y}, & x\ge 0, y \ge 0\\ 0, & \text{else}\end{array}\right. . $$ with $\alpha, \beta >0, N\in \mathbb{R}$.
For which $N$, f is a joint PDF?
Okay, I know the characteristics of a joint PDF: $$(1) \int_{\mathbb{R}^2}f(x,y)dxdy=1$$ $$(2)\quad f(x,y)\ge 0$$
The second assumption is true for $N>0$. I've tried to get the right $N$ so that the integral is $1$ but I had problems with the integration respect to $x$ and $y$.
Any hints? Thank you so much!
We define f on $\mathbb{R^2}$ such that $$f(x,y)=N e^{-\alpha x-\beta y}1_{x\geq0}1_{y\geq0}$$
$f$ must be positive, therefore we choose $N>0$.
We apply Tonelli's theorem :
$$\int_{\mathbb{R}^2}f(x,y)dxdy=N\int_{\mathbb{R}}e^{-\alpha x}1_{x\geq0}dx\int_{\mathbb{R}}e^{-\beta y}1_{y\geq0}dy$$
We define the integrale $I(u)=\int_{\mathbb{R}}e^{-u y}1_{y\geq0}dy$ where $u>0$.
Clearly, we have $$I(u)=\int_{0}^{\infty}{e^{-u y}dy}=\left[-\frac{e^{-uy}}{u}\right]_{0}^{\infty}=\frac{1}{u}$$
Using this result we have that $$\int_{\mathbb{R}^2}f(x,y)dxdy=NI(\alpha)I(\beta)=\frac{N}{\alpha \beta}$$
We want this integral to be equal to $1$, thus, $N=\alpha \beta$