Let $X$ and $Y$ be i.i.d. random variables with Exponential$(\lambda)$ distribution. Using conditioning, determine a) $P(X < Y)$ and b) the distribution function of $X + Y$.
Attempt:
I thought I'd calculate the probability with $P(X < Y) = P(X < y |Y = y)$. To do that you first determine the conditional density, then calculate $P(X < y |Y = y)$. So,
Since $X$ and $Y$ are independent, then $f_{X, Y}(x, y) = f_X(x)f_Y(y)$. Then the conditional distribution is
$$ f_{X|Y}(x|y) = \frac{f_X(x)f_Y(y)}{f_X(x)}, ~~x>0,~~y>0 $$
That is
$$ f_{X|Y}(x|y) = f_X(x),~~x>0,~~~y>0 $$
By this point I'm already thinking I'm doing it totally wrong because when is a problem that easy, right? Anyway, continuing this line of reasoning you get
$$ P(X < x|Y = y) = \int_{0}^{y}\lambda e^{-\lambda x}dx = F_X(y) = 1 - e^{-\lambda y}$$
Official Answer: $P(X < Y) = 1/2 $.
Since my answer for a) is wrong, I didn't even try b).
We have \begin{align} \mathbb P(X<Y) &= \int_0^\infty \mathbb P(X<Y\mid Y=y) f_Y(y)\ \mathsf dy\\ &=\int_0^\infty \int_0^y f_X(x)\ \mathsf dx\ f_Y(y)\ \mathsf dy\\ &=\int_0^\infty \int_0^y \lambda e^{-\lambda x}\ \mathsf dx \lambda e^{-\lambda y}\ \mathsf dy\\ &=\int_0^\infty (1-e^{-\lambda y})\lambda e^{-\lambda y}\ \mathsf dy\\ &=\frac12. \end{align}
For the second part, convolution is the simplest method; for $t\geqslant 0$ the density of $Z:=X+Y$ is \begin{align} f_Z(t) &= (f_X\star f_Y)(t)\\ &= \int_{\mathbb R}f_X(s)f_Y(t-s)\ \mathsf ds\\ &= \int_0^t \lambda e^{-\lambda s}\lambda e^{-\lambda(t-s)}\ \mathsf ds\\ &= \lambda^2\lambda e^{-\lambda t}\int_0^t \ \mathsf ds\\ &= \lambda(\lambda t)e^{-\lambda t}. \end{align} I suppose there is a way to obtain this by conditioning but the computations would likely be more complicated.