Find expected value for $\max (X_1,...,X_n)$

Question

I have a random variable $X$ behave as $ U[0,\alpha]$.

Out of $n$ experiments, I would look at the new random variable $Y=\max(X_1,...X_n)$.

I ask for $E(Y)$.

I know it supposed to be $\alpha \cdot \frac{n-1}{n}$ or something like that (maybe $\alpha \cdot \frac {n}{n+1}$) and I understand why, but all this is intuitive and I don't understand how to prove it.

Thanks in advance for your answers

Answer 1

Let $$Z = max(X_1,X_2,...X_n)$$

$$P(Z\le z) = P(max(X_1,X_2,...X_n)\le z)$$

$$= F(X\le z)^{n}$$

$$ = (\frac{z}{\alpha})^{n}$$

pdf is the derivative of it

Hence $$f_Z(z) = \frac{n.z^{n-1}}{\alpha^n}$$

$$E(Z) = \int_{0}^{\alpha} \frac{nz..z^{n-1}}{\alpha^n} = \frac{n\alpha}{(n+1)}$$

Answer 2

$$\mathbb P\{Y\leq y\}=\mathbb P\{X_1\leq y,...,X_n\leq y\}=\mathbb P\{X_1\leq y\}^n=F_{X_1}(y)^n.$$ Therefore $$f_Y(y)=nf_{X_1}(y)F_{X_1}(y)^{n-1}=n\frac{1}{\alpha^n }y^{n-1}\boldsymbol 1_{[0,\alpha ]}(y).$$

Finally, $$\mathbb E[Y]=\int_0^\alpha yf_Y(y)\,\mathrm d x=...$$