$X_1$ and $X_2$ are independent and identically distributed (i.i.d) random variables defined on R+ each with pdf of the form $f_X(x) = \sqrt\frac{1}{2\pi x}exp[\frac{-x}{2}]\quad ,\quad x>0, \quad 0 \quad otherwise.\quad$ I need to find the joint distribution of random variables $Y_1 = X_1$ and $Y_2 = X_1 + X_2.$

I will need to find the pdf of $Y_2$ using convolution and afterwards I can just multiply the result with $Y_1$ to get the joint distribution. Where I have been able to get up to is as follows:

$f_{Y_2}(y_2) = \int^\infty_{-\infty} f_{X_1}(y_2-x_2)f_{X_2}(x_2)\;dx_2$

$f_{Y_2}(y_2) =\int^{y_2}_{0}\sqrt\frac{1}{2\pi (y_2-x_2)}e^\frac{-(y_2-x_2)}{2}\sqrt\frac{1}{2\pi x_2}e^\frac{-x_2}{2}\;dx_2$

$f_{Y_2}(y_2) =\frac{e^\frac{-y_2}{2}}{2\pi }\int^{y_2}_{0}(x_2y_2-{x_2^2})^\frac{-1}{2} \:dx_2$

Trying to evaluate this integral is giving me trouble and after using wolfram alpha it's suggesting a division by zero so I'm feeling like I've missed a crucial step / trick in this process - have I taken the right approach and how could I proceed from here?

  • 1
    $\begingroup$ What do you mean "just multiply the result with $Y_1$ to get the joint distribution"? $Y_1$ and $Y_2$ are not independent, so you can't just multiply the distributions to get the joint distribution, if that's what you're thinking. $\endgroup$
    – jbowman
    Commented Jul 31, 2018 at 15:20
  • $\begingroup$ Yes - I only looked at the fact that both $X_1$ and $X_2$ are i.i.d but didnt realise that since both $Y_1$ and $Y_2$ consist of $X_1$, they are related and not independant. Thanks for pointing this out $\endgroup$
    – Maharero
    Commented Jul 31, 2018 at 21:45

2 Answers 2


This problem is easier than you are making it out to be; there's no need for integration.

First, we write out the joint distribution of $X_1$ and $X_2$:

$$f(x_1,x_2) = {1\over 2\pi}(x_1x_2)^{-1/2}e^{-{1\over 2}(x_1+x_2)}$$

We make the transformations $y_1 = x_1$ and $y_2 = x_1+x_2$, noting that the Jacobian of the transforms is 1, so we can ignore it in the subsequent steps. Substituting $y_1$ for $x_1$ and $y_2-y_1$ for $x_2$ everywhere gives:

$$f(y_1,y_2) = {1\over 2\pi}(y_1(y_2-y_1))^{-1/2}e^{-{1\over 2}y_2}1_{y_2>y_1}$$

where $1_a$ is the indicator function that takes the value $1$ when $a$ is true and $0$ otherwise.


Try to use Integral Calculator. Here what I got:

$\int (x_2 y_2 - x_2^2) ^ {-\frac{1}{2}} dx_2= \arcsin\left(\dfrac{2x_2-y_2}{y_2}\right) + C$

Add close interval

$\int_{0}^{y_2} (x_2 y_2 - x_2^2) ^ {-\frac{1}{2}} dx_2 = \arcsin\left(\dfrac{2x_2-y_2}{y_2}\right) |_0^{y_2} = \arcsin(1) - \arcsin(-1) = \pi$

To sum up $f_{Y_2}(y_2) = \frac{1}{2} e^{\frac{-y_2}{2}}$

  • $\begingroup$ Thanks for showing me this website - best thing about it is that it shows all the steps involved so you can learn from it (I didnt consider completing the square first). My initial assumption of using this answer to create the joint distribution wont work due to dependance like jbowman pointed out but this is good to know for the future! $\endgroup$
    – Maharero
    Commented Jul 31, 2018 at 22:16

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