# Convolution of a less typical distribution

$X_1$ 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?

• 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. Commented Jul 31, 2018 at 15:20
• 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 Commented Jul 31, 2018 at 21:45

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.

• I wasnt aware of the transformation approach - after reading through it here (markirwin.net/stat110/Lecture/Section36.pdf) I now completely follow your answer. Thank you! Commented Jul 31, 2018 at 21:30

Try to use Integral Calculator. Here what I got:

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

$\int_{0}^{y_2} (x_2 y_2 - x_2^2) ^ {-\frac{1}{2}} dx_2 = \arcsin\left(\dfrac{2x_2-y_2}{y_2}\right) |_0^{y_2} = \arcsin(1) - \arcsin(-1) = \pi$
To sum up $f_{Y_2}(y_2) = \frac{1}{2} e^{\frac{-y_2}{2}}$