# Suppose $Z_i$ are i.i.d. $N(0, 1).$ And let $M_n = \max\{Z_1, \ldots, Z_n\}$, show that $P(M_n > t) \leq n(1 - \Phi(t))$

Show that $$P(M_n > t) \leq n(1 - \Phi(t))$$

My work:

\begin{align} & P(M_n > t) \leq P\left(\bigcup_{i=1}^n (Z_i > t) \right) \\ \leq {} & \sum_{i=1}^n P(Z_i > t)= n(1 - \Phi(t)) \end{align}

Is the above correct?

• A first step is showing $P(M_n > t) =1-\Phi (t)^n$.
– JimB
Commented Aug 2 at 3:12
• @JimB Why do the long way?
– BCLC
Commented Aug 7 at 18:33
• @BCLC Nice (and short) answer. But I assume (or at least hope) you don't believe that there's only one way to do something. Plus, because the right-hand side of the equation was targeted for the normal distribution, it would seem appropriate that the left-hand side of the equation should also be specific to the normal distribution.
– JimB
Commented 2 days ago

\begin{align} & P(M_n > t) \leq P\left(\bigcup_{i=1}^n (Z_i > t) \right) \\ \leq {} & \sum_{i=1}^n P(Z_i > t)= n(1 - \Phi(t)). \end{align}

The above is correct but I would write it like this: \begin{align} & P(M_n > t) = P\left(\bigcup_{i=1}^n (Z_i > t) \right) \\ \leq {} & \sum_{i=1}^n P(Z_i > t)= n(1 - \Phi(t)). \end{align}

When I read the question I spent a few seconds on whether it ought to say $$\text{“}{\le}{”}$$ or $$\text{“}{\ge}{”}.$$ It seems clearer if one says that $$\left\{ M_n>t \right\}$$ and $$\left\{ \bigcup_{i\,=\,1}^n (Z_i>t) \right\}$$ are not just events with the same probability, but are the same event.