$X$ and $Y$ are independent random variables,then density function of $\frac{X+Y}{3} $? $X$ and $Y$ are independent random variables each having the density $f(t) = \frac{1}{\pi(1+t^2)} , -\infty < t < \infty$.
Then the density function of $\frac{X+Y}{3}$ is?
Let $Y =U$ and  $X+Y = V$ 
then I was thinking of calculating the Joint distribution of $ U , V $ and then calculating the marginal density function $f_{V}$!
So $f(u,v) = |J| . f(x,y)$
where the Jacobian is $-3$ hence $|J| = 3.$
So that $f(u,v) = 3 . f(x,y) = 3. f(x) . f(y) = 3 . \frac{1}{\pi (1 + x^2)}\frac{1}{\pi (1 + y^2)} =3 . \frac{1}{\pi (1 + (3v - u)^2)}\frac{1}{\pi (1 + u^2)}  $
And finally $f(v) = \int_{-\infty}^{\infty} \frac{3}{\pi^2 (1 + u^2)(1 + (3v - u)^2)} du$
Now, are my above steps correct?.
Also, it is becoming a bit complicated any other simpler method?
EDIT
It's a problem from  previous year competition in our institute which had four options as answer - 
$a) \frac{6}{\pi (4 + 9t^2)}$
$b) \frac{6}{\pi (9+4t^2)}$
$c) \frac{3}{\pi (1 + 9t^2)}$
$d) \frac{3}{\pi (9 + t^2)}$
and usually there is a small amount of time for each question any reverse mathematics procedure or any tricks may be there apart from a few lengthy calculations, yes from answers I now know that characteristic functions are important to know about various properties of distributions but to a beginner in probability if unaware of  Cauchy Random variable?
 A: Characteristic functions provide an efficient way to compute properties of distributions. 
Here, $X$ and $Y$ are Cauchy random variables (which you can tell from the form of the PDF). The characteristic function of a Cauchy random variable is in general:
$$f(x) = \frac{1}{\pi \gamma \left[1+\left(\frac{x-x_0}{\gamma}\right)^2\right]} \quad\leftrightarrow\quad \varphi(t) = \exp{\left(x_0 it - \gamma |t|\right)}$$
And here specifically for $X$ and $Y$, we can see that:
$$\varphi_X(u) = \varphi_Y(u) =  \exp{(-|u|)}$$

Next, characteristic functions combine nicely under linear combinations of independent random variables:

Put $Z \equiv \frac{X+Y}{3}$. Then the characteristic function of $Z$ is:
$$\begin{align*}\varphi_Z(u) &=  \varphi_X(u/3) \cdot \varphi_Y(u/3)\\
&= \exp{(-|u/3|)} \cdot \exp{(-|u/3|)}\\
&= \exp{(-2|u/3|)}\\
&= \exp{\left(-\frac{2}{3}|u|\right)}
\end{align*}$$
And I claim that this has the form of a Cauchy distribution; we can use the "density-to-characteristic-function" conversion above to back-solve for the parameters: $\gamma=2/3$, $x_0 = 0$.
Hence $Z = (X+Y) / 3$ is a Cauchy random variable with $\gamma = 2/3$ and $x_0 = 0$; its density function is
$$f_Z(t) = \frac{1}{\frac{2}{3}\pi \left[1+(3t/2)^2\right]}$$
A: First, we can find a distribution function $F_{(X+Y)/3}$ 
$$
F_{(X+Y)/3}(x) = \mathbb{P}\left\{\frac{X+Y}{3} \leq x\right\} = \int \mathbb{P}\left\{\frac{X+y}{3} \leq x \; \Bigg| \; Y = y\right\} \; d\mathbb{P}\{Y \leq y\} = \\[7pt] = \int \mathbb{P}\left\{X \leq 3x - y \right\} f(y) \, dy = \int\limits_{-\infty}^\infty dy \, f(y) \int\limits^{3x-y}_{-\infty} \, dx \, f(x).
$$
Density is given by $f_{(X+Y)/3}(x) = F_{(X+Y)/3}'(x)$:
$$
f_{(X+Y)/3}(x) = 3\int\limits_{-\infty}^\infty dy \, f(y) f(3x-y).
$$
