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$X_1, X_2, \ldots, X_n$ are $n$ i.i.d. uniform random variables. Let $Y = \min(X_1, X_2,\ldots, X_n)$. Then, what's the expectation of $Y$(i.e., $E(Y)$)?

I have conducted some simulations by Matlab, and the results show that $E(Y)$ may equal to $\frac{1}{n+1}$. Can anyone give a rigorous proof or some hints? Thanks!

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  • $\begingroup$ This is known as the $Beta(1,n+1)$ distribution. $\endgroup$ Commented Mar 10, 2023 at 22:25

3 Answers 3

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To calculate the expected value, we're going to need the density function for $Y$. To get that, we're going to need the distribution function for $Y$. Let's start there.

By definition, $F(y) = P(Y \leq y) = 1 - P(Y > y) = 1 - P(\text{min}(X_1, \ldots, X_n) > y)$. Of course, $\text{min}(X_1, \ldots X_n) > y$ exactly when $X_i > y$ for all $i$. Since these variables are i.i.d., we have $F(y) = 1 - P(X_1 > y)P(X_2>y)\ldots P(X_n>y) = 1 - P(X_1 > y)^n$. Assuming the $X_i$ are uniformly distributed on $(a, b)$, this yields $$F(y) = \left\{ \begin{array}{ll} 1 - \left(\frac{b-y}{b-a}\right)^n & : y \in (a, b)\\ 0 & : y < a\\ 1 & : y > b \end{array} \right.$$ We take the derivative to get the density function. $$f(y) = \left\{ \begin{array}{ll} \frac{n}{b-a} \left(\frac{b-y}{b-a}\right)^{n-1} & : y \in (a, b)\\ 0 & : \text{otherwise} \end{array} \right.$$ Now $E(Y) = \int_{-\infty}^{\infty} y f(y) dy$. The integral is straightforward; I'll leave the details to you. I calculate $E(Y) = \frac{b+na}{n+1}$.

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  • $\begingroup$ Thanks for your answer. You extend it to a more general setting, which is very helpful. $\endgroup$
    – jet
    Commented May 8, 2014 at 13:42
  • $\begingroup$ There's no sense in taking the derivative and then integrating; you can just use the tail-sum formula to calculate the expectation directly from the integral. $\endgroup$
    – Zaz
    Commented Apr 25, 2022 at 23:36
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Yes. Assuming a $U(0,1)$, note that

$$\Pr\Bigl(\min_i X_i\leq x\Bigr) = 1- \Pr\Bigl(\min_i X_i\geq x\Bigr) = 1- (1-x)^n. $$

So the density function is

$$f(x)=n(1-x)^{n-1}.$$

Then

$$\int_0^1 x f(x) dx = n\int_0^1 x(1-x)^{n-1} dx = n\int_0^1 (1-t) t^{n-1} dt = \frac{1}{n+1}. $$

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    $\begingroup$ Thanks for your answer, which is very helpful. $\endgroup$
    – jet
    Commented May 8, 2014 at 13:43
  • $\begingroup$ Jpi what if $X\sim U(0,\theta)$ ? I am not able to solve the integral when I am trying to find expectation . @jet Can you answer my question ? $\endgroup$
    – Daman
    Commented Oct 12, 2018 at 16:36
  • $\begingroup$ @Daman divide the $X_i$'s by $\theta$ first, which is equivalent to multiplying $x$ by $\theta$ in the first display. Hth $\endgroup$
    – JPi
    Commented Sep 3, 2019 at 23:59
  • $\begingroup$ the change of variable from x to t would require (-dt) and switching the boundary of integral from 1 to 0 - but both cancel out. i.e. $$ n\int_0^1 x(1-x)^{n-1} dx = n\int_1^0 (1-t) t^{n-1} (-dt) = n\int_0^1 (1-t) t^{n-1} dt. $$ $\endgroup$ Commented Feb 1, 2020 at 14:33
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An short proof without the use of the pdf. Suppose $X_i \sim U(0,1)$. Then \begin{align*} E[\min(X_i)] &= \int_0^1 P[\min(X_i)>\mu] d\mu \\ &= \int_0^1 \prod_i P[X_i>\mu] d\mu \\ &= \int_0^1 (1-\mu)^n d\mu\\ &= \int_0^1 \mu^n d\mu\\ &= \dfrac{1}{n+1} \end{align*}

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  • $\begingroup$ How does the 4th integral equals to the 3rd one? $\endgroup$
    – Our
    Commented Jan 21 at 15:31
  • $\begingroup$ @Our Change of variables $\mu'=1-\mu$. $\endgroup$ Commented Jan 31 at 16:42

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