Archive for January, 2017

Expected value of minimum of n i.i.d U(0,1) random variables

January 23, 2017

What is the expected value of minimum of n i.i.d random variables that are drawn from U(0,1) distribution

Let u_1, u_2, ... , u_n are the n random variables.

Let X = min(u_1, u_2, ... , u_n)

Plan of Action

  1. Find CDF for X
  2. Find PDF for X using CDF
  3. Compute E[X] = \int_{0}^{\infty} x\, f_X(x) \,dx

\begin{array}{rcl} F_X(x) &=& P(X \leq x) \\ &=& 1-P(X>x) \\ &=&  1- [ P(u_1>x) \wedge P(u_2>x) \wedge \dots \wedge P(u_n>x) ] \\ &=& 1-(1-x)^n \end{array}

Note that F_X(x) = \int_{-\infty}^{x} f_X(t) \,dt

Thus f_X(x) = \frac{d}{dx}F_X(x)

so f_X(x) = n (1-x)^{n-1}

\begin{array}{rcl} E[X] &=& \int_{0}^{\infty} x f_X(x) \,dx \\ &=&n \,\int_{0}^{1} x\,(1-x)^{n-1}\, dx \\ &=&\frac{1}{n+1} \end{array}

Alternative way :

For non-negative random variables, some times it is easy to calculate expected value of random variable using following formula :

E[X] = \int_{0}^{\infty} P(X>x) \,dx

\begin{array}{rcl} P(X>x) &=& P(u_1>x) \wedge P(u_2>x) \wedge \dots \wedge P(u_n>x) \\ &=&  (1-x)^n \end{array}

Since in our case X cannot take value greater than 1

E[X] = \int_{0}^{1} P(X>x) \, dx =\int_{0}^{1}(1-x)^n \,dx = \frac{1}{n+1}

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