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Chapter 2: Problem 48

If \(X\) is uniformly distributed over \((0,1)\), calculate \(E\left[X^{2}\right]\).

Short Answer

Expert verified
The expected value of \(X^2\) when \(X\) is uniformly distributed over \((0,1)\) is \(\frac{1}{3}\).

Step by step solution

01

Write down the formula for expected value

For a continuous random variable \(X\) with probability density function \(f(x)\), the expected value of \(g(X)\) is given by: \(E[g(X)] = \int_{-\infty}^{\infty} g(x)f(x) dx\) In our case, we are interested in finding \(E[X^2]\), so we will use \(g(X) = X^2\).
02

Set up the integral for \(E[X^2]\)

Since \(X\) is uniformly distributed over \((0,1)\), we know that the integral for the expected value will be computed over this interval. Therefore, we can set up the integral as follows: \(E[X^2] = \int_{0}^{1} x^2 f(x) dx\) Recall that the probability density function of a uniform distribution over \((0,1)\) is \(f(x) = 1\). Thus, we have: \(E[X^2] = \int_{0}^{1} x^2 dx\)
03

Compute the integral

Now, we need to compute the integral: \(E[X^2] = \int_{0}^{1} x^2 dx\) We know that \(\int x^2 dx = \frac{1}{3}x^3 + C\). Using the fundamental theorem of calculus, we get: \(E[X^2] = \left[\frac{1}{3}x^3\right]_0^1 = \frac{1}{3}(1)^3 - \frac{1}{3}(0)^3 = \frac{1}{3}\)
04

Interpret the result

The expected value of \(X^2\) when \(X\) is uniformly distributed over \((0,1)\) is \(\frac{1}{3}\). This means that, on average, the value of \(X^2\) will be \(\frac{1}{3}\) when \(X\) follows this distribution.

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