1
$\begingroup$

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!

1 Answers 1

1

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, $N=\alpha \beta$

  • 1
    It is a consequence of the Tonelli's theorem, you can write $$\int_{\mathbb{R}^2}f(x,y)dxdy=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}f(x,y)dx\right)dy=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}f(x,y)dy\right)dx$$ – 2017-01-09
  • 0
    Thanks! One additional question: How I can evaluate $\mathbb{E}[X]$ without $Y$, if I have the distribution function? – 2017-01-09
  • 1
    $$E(X)=\int_{\mathbb{R}^2}xf(x,y)dxdy=\int_{\mathbb{R}}x \left(\int_{\mathbb{R}}f(x,y)dy\right)dx$$. You must now calculate $$\int_{\mathbb{R}}f(x,y)dy$$ – 2017-01-09