Revision as of 07:08, 25 February 2013

Practice Problem: normalizing the probability mass function of a continuous random variable

A random variable X has the following probability density function:

$ f_X (x) = C e^{\frac{-x^2}{2}} , $

where C is a constant. Find the value of the constant C.

Share your answers below

You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!

Answer 1

By the properties of pdfs we know that:

$ \int\limits_{-\infty}^{+\infty} f_X(x) dx = 1. $

$ Let, I = \int\limits_{-\infty}^{+\infty}e^{\frac{-x^2}{2}}dx. $

By introduction a new term $ e^{\frac{-y^2}{2}} $ we know that:

$ I^2 = \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}e^{\frac{-x^2}{2}}e^{\frac{-y^2}{2}}dxdy = \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}e^{\frac{-x^2-y^2}{2}}dxdy = \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}e^{-\frac{1}{2}\left(x^2 + y^2\right)}dxdy. $

Now, we need to change to polar coordinates. We know that $ r^2 = x^2 + y^2, $ and that the Jacobian for Polar coordinates is $ r. $ By performing the substitution we have:

$ I^2 = \int\limits_{0}^{+\infty}\int\limits_0^{2\pi}re^{-\frac{1}{2}r^2}d{\theta}dr. $

Now, we just need to evaluate the integral:

$ I^2 = \int\limits_{0}^{+\infty}\int\limits_0^{2\pi}re^{-\frac{1}{2}r^2}d{\theta}dr = 2\pi\int\limits_{0}^{+\infty}re^{-\frac{1}{2}r^2}dr = 2\pi\left(-e^{-\frac{1}{2}r^2}\right)\bigg|_{0}^{+\infty} = 2\pi\left(0 + 1\right) = 2\pi. $

Now, we solve for $ I $:

$ I = \sqrt{2\pi}. $

Returning to the original problem, we actually want to solve:

$ 1 = \int\limits_{-\infty}^{+\infty} f_X(x) dx = \int\limits_{-\infty}^{+\infty}C e^{\frac{-x^2}{2}}dx = C\int\limits_{-\infty}^{+\infty}e^{\frac{-x^2}{2}}dx = C \times I = C \times \sqrt{2\pi}. $

Now, solving for the constant $ C: $

$ C = \frac{1}{\sqrt{2\pi}}. $

• Instructor's comments: Good job! You have a good approach, and all your steps are clearly explained. Now I want to challenge the other students: can you write a shorter but equivalent solution? In order words, can you "compactify" the justification? -pm

Answer 2

Since $ f_X (x) $ is in the form of a 1-D gaussian random variable, whose PDF is of the form $ \frac{1}{\sqrt{2\pi}s}e^{-\frac{(x-m)^2}{2s^2}} $, it can be seen from the problem statement that m is zero and $ s^2 $ is one. Therefore, the value of C is $ \frac{1}{\sqrt{2\pi}s} $, which is $ \frac{1}{\sqrt{2\pi}} $, since s is either positive or negative one (and only positive one will yield a logical answer).

Answer 3

Write it here.

Back to ECE302 Spring 2013 Prof. Boutin

Back to ECE302