Q:

show that $P(Y_{(n)} > m) = 1 − (.5)^n$ .

If Y is a continuous random variable and m is the median of the distribution, then m is such
that $P(Y ≤ m) = P(Y ≥ m) = 1/2$. If $Y_1, Y_2, . . . , Y_n$ are independent, exponentially distributed random variables with mean β and median m, $U = Y_{(n)}$ =
max($Y_1, Y_2, . . . , Y_n$) does not have an exponential distribution. Use the general form of$ F_U (y)$ to show that $P(Y_{(n)} > m) = 1 − (.5)^n$ .
Could someone get me started on this?

A:

I calculate firstly the probability distribution function of $U=Y_{(n)}$. It is $$F_{U}(u)= P(Y_{(n)}\le u)=P(\max\{Y_1,\ldots,Y_n\le u\})=P(Y_1\le u,\ldots,Y_n\le u)=^{independent}=P(Y_1\le u)\cdot\ldots\cdot P(Y_n\le u)=^{equally \,distributed}(P(Y\le u))^n=(F_Y(u))^n$$ $%=^{exponential \, distribution}(1-\exp(-\beta y))^n.$
So we have: $$P(Y_{(n)}>m)=P(U>m)=1-P(U\le m)=1-F_U(m)=1-(F_Y(m))^n=1-0.5^n$$
where the last equation holds because $F_Y(m)=P(Y\le m)=0.5$ according to the definition of the median. 

