I have a relatively simple question regarding a joint PDF given to me.

There are 2 random variables X and Y, with the following joint PDF. PDF

a) Find the value of a.

I have attempted to set up the following integral: $$ a\int_{0}^{\infty }\int_{-\infty}^{\infty } e^{-x-2y}dydx = 1 $$

However, this integral diverges to infinity, so no value of a works. I'm guessing I'm missing some fundamental property of a joint PDF - please fill me in!


  • 1
    $\begingroup$ if $y<0$ then $f(x,y)=0$ so $\int_{-\infty}^{\infty}$ should be changed into $\int_0^{\infty}$. $\endgroup$
    – drhab
    Commented Apr 13, 2015 at 9:58
  • $\begingroup$ Why is y < 0 mean the joint pdf is 0? $\endgroup$
    – Kevin
    Commented Apr 13, 2015 at 11:01
  • $\begingroup$ By definition: $0$ elsewhere... $\endgroup$
    – drhab
    Commented Apr 13, 2015 at 11:17
  • $\begingroup$ I think you misunderstood the use of the comma. $0<x,y<1$ mean both variables are in $(0,1)$, for example. $\endgroup$
    – jdods
    Commented Nov 6, 2017 at 1:35

2 Answers 2


Because the joint PDF of $X, Y$ is just the combination of independent exponential random variables, we could find the normalized constant $a=2$ easily from the definition of exponential random variables $f_X(x)=\lambda exp (-\lambda x)$.

$f_X,_Y(x,y)=f_X(x) f_Y(y)=exp(-x)\cdot 2 exp(-2y).$


The integral should be set:

$$\int_{0}^{\infty}\int_{0}^\infty ae^{-x-2y}dydx=1$$

where the boundaries are $0<x<\infty$ and $0<y<\infty$

Hope this helps.


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