Simple question on differentiation of cdf [duplicate]

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 $$y\,?$$.
• Are the $X_i's$ independent? – Ekesh Kumar Nov 11 '20 at 15:30
• @EkeshKumar yes – historgn Nov 11 '20 at 15:33
• Okay. See my answer. – Ekesh Kumar Nov 11 '20 at 15:33
• – StubbornAtom Nov 11 '20 at 16:49

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:

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

• in your definition of $f_Y(y)$ there is an error. $f_Y(y)=0$ when $y>1$... you wrote $f_Y(y)=1$ – tommik Nov 11 '20 at 15:47
• @tommik Thank you for catching that. I have corrected it. – Ekesh Kumar Nov 11 '20 at 15:47

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

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