# If $X_1$ and $X_2$ are continuous random variables from a $U(0,10)$ distribution, then what is the probability density function of $X_1 + X_2$?

Suppose that $$X_1, X_2 \sim U(0,10)$$ are continuous random variables. How would I derive the probability density function of $$X_1 + X_2$$? Clearly we have $$f_{X_1}(x) = f_{X_2} = 0.1$$ and so $$f_{X_1+X_2}(x) = \int_{-\infty}^\infty f_{X_1}(s) \hspace{1mm} f_{X_2}(x-s) \hspace{2mm} ds$$ (if anyone knows what this rule is called, if it has a name, please let me know).

Now, since $$f_{X_1}(x) = 0.1$$ if $$0 \leq x \leq 10$$ and is $$0$$ otherwise, this integral can be simplified to $$f_{X_1+X_2}(x) = 0.1 \int_0^{10} f_{X_2}(x-s) \hspace{2mm} ds$$

This is where I struggle to see what should happen next. Would anyone be able to help me to understand how to proceed from here?

Cheers.

• Mar 18, 2020 at 15:30
• It sounds like you are talking about "convolution," which is only applicable when $X_1$ and $X_2$ are independent. In your final integral I observe that $f_{X_2}(x-s)$ is only nonzero if $x-s \in [0,10]$, that is, $0\leq x-s \leq 10$. Mar 18, 2020 at 15:52
• What you need to do now is to calculate the integral of a piecewise function - there's no stochastics involved at this point anymore. I.e. define $$f(x):= f_{X_2}(x) = \begin{cases} 0 &, x<0\\ 0.1 &,x\in [0,10]\\ 0 &, x>10\end{cases}$$ and calculate $\int_0^{10} f(x-s)ds$ Mar 18, 2020 at 16:42
• Are $X_1$ and $X_2$ independent? Mar 18, 2020 at 17:57

$$Y=X_1+X_2$$
$$f_Y(y)=0.1 \int_0^{10} f_{X_2}(y-s) ds$$
we need to clarify the area $$0
Split in two situation $$y<10$$ and $$10 $$\begin{eqnarray} f_Y(y)&=&\left\{ \begin{array}{cc} 0.1 \int_0^{y} f_{X_2}(y-s) ds & y\leq 10 \\ 0.1 \int_{y-10}^{10} f_{X_2}(y-s) ds & 10