The random variables $X_1, X_2,...$ are independent and $\epsilon \ N(0,1)$. Set
$N = \min\{k \mid X_k \geq 0\}$
Show that the pdf of $X_N$ is
$f_{X_N}(x) = 2\phi(x)$ for $x \geq 0, 0$ elsewhere
Aid: it may be helpful to compute the probability $P(X_N \leq x |N = n)$
The solution is:
$P(X_N \leq x | N = n) = P(X_n \leq x | X_1 \le 0 ,..., X_n \geq 0)$
by independence we get
$ = P(X_n \leq x | X_n \geq 0)$
and as r.v.s are I.I.D
$ = P(X_1 \leq y | X_1 \geq 0)$
and by defintion of conditional probability
$ = \frac{P(0 \leq X_1 \leq y)}{P(X_1 \geq 0)}$
$ = \frac{\phi(x)-\phi(0)}{1/2}$ as $x \geq 0$
$ = 2\cdot (\phi(x) -(\phi(0)) $
Thus $X_N$ is independent of N and we get
$f_{X_N}(x) = \frac{d}{dx}P(X_N \leq x|N = n) = \frac{d}{dx}P(X_N \leq x) = 2\cdot \phi(x)I(x\geq0)$
I wonder the following:
How does $N = \min\{k \mid X_k \geq 0\}$ affect anything ?
And how do we get from
$ = \frac{P(0 \leq X_1 \leq y)}{P(X_1 \geq 0)}$
to
$ = \frac{\phi(x)-\phi(0)}{1/2}$ as $x \geq 0$