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Suppose we have $n$ $i.i.d$ random variables $X_{1},\ldots,X_{n}$ all distributed uniformly, $X_{i} \sim \mathrm{Uniform}\left(0,1\right)$ .

  • We want to find the expected value of $\mathbb{E}[Y_{n}]$ where $Y_{n} = \max\left\{X_{1},\ldots,X_{n}\right\}$.
  • Here I got $F_Y\left(y\right) = \int\mathrm{d}x_{1}\ldots\int\mathrm{d}x_{n} = \left[F_{X}\left(y\right)\right]^n $ but to get $f_{Y}\left(y\right)$ i need to differentiate $F_{Y}\left(y\right)$ with respect to $y$ but how do i differentiate $\left[F_{X}\left(y\right)\right]^{n}$ with respect to $y\,?$.
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Define $Y = \max_{1 \leq i \leq n} X_i$ so that for any $0 \leq y \leq 1$, we have

$$F_{Y}(y) = P(Y \leq y ) = \prod_{i = 1}^{n} P(X_i \leq y) = [P(X_1 \leq y)]^{n} = y^{n}.$$

Differentiating, one can find

$$f_{Y}(y) = \frac{d}{dy} F_{Y}(y) = ny^{n - 1},$$

which yields

$$f_{Y}(y) = \begin{cases} ny^{n - 1} & \text{ if } 0 \leq y < 1 \\ 0 & \text{ otherwise.} \end{cases} $$

The expected value is computed as follows:

$$\mathbb{E}[Y] = \int_{0}^{1} y \cdot f_{Y}(y) \mathop{dy} = \int_{0}^{1} ny^{n} \mathop{dy} = \boxed{\frac{n}{ n + 1}}$$

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  • $\begingroup$ in your definition of $f_Y(y)$ there is an error. $f_Y(y)=0$ when $y>1$... you wrote $f_Y(y)=1$ $\endgroup$ – tommik Nov 11 '20 at 15:47
  • $\begingroup$ @tommik Thank you for catching that. I have corrected it. $\endgroup$ – Ekesh Kumar Nov 11 '20 at 15:47
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Differentiation of $F(y)$ is not necessary.

In fact,

$$F_Y(y)=y^n$$

But as per the fact that Y is non negative, its expectation can be derived in the following way

$$\mathbb{E}[Y]=\int_0^1 [1-y^n]dy=\frac{n}{n+1}$$

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