Max of n normal random variates

Question

Suppose $X_1, X_2, \ldots, X_n$ are iid distributed as $N(0,1)$. Define: $$ Y=1+\max _i\left|X_i\right| $$ I want the distribution (CDF) of $Y$.
My attempt: Let $V=\max _i\left|X_i\right|$ Since $X_i \sim N(0,1)$, the distribution of $X_i$ $$ \Phi(x)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^x e^{-t^2 / 2} d t $$ Also, the CDF of $|X_i|$ is given by $$ \begin{aligned} F_{|X_i|}(x) & =\Phi(x)-\phi(-x), x>0 \\ & =2 \times \frac{1}{\sqrt{2 \pi}} \int_0^x e^{-t^2 / 2} d t \\ & =\frac{1}{\sqrt{2}} \times \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2 / 2} d t \\ & =\frac{1}{\sqrt{2}} \times \operatorname{erf}(x) \end{aligned} $$ $\therefore$ CDF of $Y$ is given by: $$ F_Y(y)=\left(\frac{1}{\sqrt{2}} erf (y-1)\right)^n $$ Am I right in the above calculations?Thank you for any help

Answer 1

Let $U = g(X) = |X|$. To calculate the pdf distribution of $U$, notice that $g^{-1}(U) = \pm X$. The transformation from $X$ to $U$ is given by

\begin{equation} F_U(u) = P(U\leq u)=P(g(X)\leq u) = P(X \leq \pm u) = F_X(u) - F_X(-u) \end{equation}

Now, the maximum is given by given by

\begin{equation} F_V(u) = \max(U_i)=\left(F_X(u) - F_X(-u) \right)^n, \quad u\geq 0 \end{equation}

which is where you seem to have gone wrong. Continuing, the cdf of $V$ is explicity

\begin{equation} F_V(u) = \left( \frac{1}{2}\left(1+\operatorname{erf}\left(\frac{u}{\sqrt{2}}\right)\right) - \frac{1}{2}\left(1+\operatorname{erf}\left(\frac{-u}{\sqrt{2}}\right)\right) \right)^n = \left(\operatorname{erf}\left(\frac{u}{\sqrt{2}} \right) \right)^n \end{equation}

whereafter the cdf of $Y=V+1$ is given by

\begin{equation} F_Y(y) = \left(\operatorname{erf}\left(\frac{y-1}{\sqrt{2}} \right) \right)^n, \quad y\geq 1 \end{equation}

I guess at the end, it was only your normalization factor which was wrong?

An example with $n=5$ is given below, where I have used Monte Carlo simulations to compare the empirical cdf with the analytical cdf of $V$.