$X,Y$ are independent random variables with uniform distribution on $[0,1]$, and let the random variable $Z=X+Y$.
The density of $Z$ is:
$$f_{X+Y}(z)=\int_0^z f_X(x)f_Y(z-x)dx$$
What is the formula for the probability $P(Z \leq m)$?
Any help would be appreciated.
Sorry, I have no attempt.
Best Answer
$\textbf{hint}$ $$ F(a) = P(X\leq a) $$ Where $F(X)$ is the cdf. The following is how to compute the cdf from a given pdf $f(x)$ $$ F(a) = \int^{a}_{x_{min}}f(x)dx $$ where the minimum of your support of the pdf is $x_{min}$ i.e for uniform random variable between 0 and 1, $x_{min}$ = 0. Can you see how you can solve your problem?
it doesn't seem right generally and by that I mean
$f_Z(z) = \begin{cases}{ z, \ \ \ \ 0\leq z < 1 \\ 2-z, \ \ \ \ 1\leq z\leq 2 \\ 0, \ \ \ \ \text{otherwise}} \end{cases} $
ps. can someone reformat this horredous expression for me :).
So can you take it from there?