Question

Let $X$be a uniform $(0,1)$random variable. Compute role="math" localid="1646717640777" $E\left[X^n\right]$by using Proposition $2.1$, and then check the result by using the definition of expectation.

Short answer

The required answer is$E\left[X^n\right]=\frac{1}{n+1}$.

Step-by-step solution

Step 1. Given Information.

Here, it is given that: $X$is a uniform$\left(0,1\right)$random variable.

Step 2. Write the probability density function of uniform distribution.

Let $X$be a random variable and follows a uniform distribution with parameters $\left(a=0\text{and}b=1\right)$.

The probability density function of uniform distribution is $f\left(x\right)=\frac{1}{b-a};a<x<b$

Step 3. Calculate EX from the expectation definition for uniform distribution Ua,b.

$$\begin{gathered} E\left(X\right)={\int }_a^{b}x\frac{1}{b-a}dx \\ ={\int }_0^{1}x\frac{1}{1-0}dx \\ ={\int }_0^{1}xdx \end{gathered}$$

Step 4. Compute EXn by using definition of expectation.

$$\begin{gathered} E\left(X^n\right)={\int }_0^{1}xdx \\ ={\left[\frac{x^{n+1}}{n+1}\right]}_0^1 \\ =\left[\frac{1^{n+1}}{n+1}-\frac{0^{n+1}}{n+!}\right] \\ =\frac{1}{n+1} \\ \therefore E\left(X^n\right)=\frac{1}{n+1} \end{gathered}$$