# Joint probability density for independent variables

Let $X_1$ and $X_2$ be two independent random variables each with probability density function $fX_i(x_i) = e^{-x_i}$, for $x_i > 0$ for $i = 1,2$

(a) Find the joint probability density function of X1 and X2.

(b) Find $P(X_1 > 1, X_2 < 1)$

(c) Find $P(X_1 + X_2 < 2)$

For (a) I found that the joint pdf = $e^{-x_1-x_2}$

For (b) I found that $P(X_1 > 1, X_2 < 1) = \int_0^1\int_1^\infty e^{-x_1-x_2} \;\mathrm{dx_1 dx_2}$ Is this correct?

Now My question is: How do I go about integrating (c)?

is it $\int_0^2\int_0^2 e^{-x_1-x_2} \;\mathrm{dx_1 dx_2}$?

• Re (b) you suggest to compute $P(X_2<1)$, not $P(X_1>1,X_2<1)$. Re (c) you suggest to compute $P(X_1<2,X_2<2)$, not $P(X_1+X_2<2)$. – Did Jun 29 '15 at 13:09

In part (c), your integration bounds aren't correct. You want $x_1+x_2<2$, but you let both $x_1$ and $x_2$ be able to take any value in the interval $[0,2]$ independently. For instance, if $x_1=x_2=1.5$, you get $x_1+x_2\not < 2$.
The outer integral looks ok, but the bounds on the inner integral need to depend on $x_2$ in such a way that $x_1+x_2<2$.
• OK, this looks more like it: $\int_0^2\int_0^{2-x_2} e^{-x_1-x_2} \;\mathrm{dx_1 dx_2}$ – Joz Jun 29 '15 at 14:12