Category Archives: Normal Distribution

Two Practice Problems on the Standard Normal Distribution

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This post presents two practice problems with calculation involving the standard normal distribution.

Problems
Let Z be a standard normal random variable.

  1. Evaluate \displaystyle E(\lvert Z \lvert)
  2. Evaluate \displaystyle E(Z^2)

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We show that \displaystyle E(\lvert Z \lvert)=\sqrt{\frac{2}{\pi}}. Problem 2 is left as an exercise.

Let X=\lvert Z \lvert. The cumulative distribution function of X is F(x)=P(X \le x). We have the following.

\displaystyle \begin{aligned}(1) \ \ \ \ \ F(x)&= P(X \le x) \\&\text{ } \\&= P(\lvert Z \lvert \le x) \\&\text{ } \\&=P(-x \le Z \le x) \\&\text{ } \\&=\int_{-x}^x \frac{1}{\sqrt{2 \pi}} \ e^{-\frac{t^2}{2}} \ dt \\&\text{ } \\&=2\int_{0}^x \frac{1}{\sqrt{2 \pi}} \ e^{-\frac{t^2}{2}} \ dt \\&\text{ } \end{aligned}

Upon differentiation of this cdf, we have the probability density function (pdf) of X.

\displaystyle (2) \ \ \ \ \ f(x)=\sqrt{\frac{2}{\pi}} \ e^{-\frac{x^2}{2}}

The following is the calculation for E(X).

\displaystyle (3) \ \ \ \ \ E(X)=\sqrt{\frac{2}{\pi}} \ \int_0^\infty \ x e^{-\frac{x^2}{2}}=\sqrt{\frac{2}{\pi}}