Latest revision as of 14:00, 25 March 2013

Practice Problem: compute the first order moment of a Gaussian random variable

A random variable X has the following probability density function:

$ f_X (x) = \frac{1}{\sqrt{2\pi} 3 } e^{\frac{-(x-3)^2}{18}} . $

Compute the moment of order one of that random variable. In other words, compute

$ E \left( X^1 \right) . $

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Answer 1

The moment of n-th order moment is defined as: $ E[X^{n}]=\int_{-\infty }^{\infty }x^{n}f_{X}(x)dx $

Therefore,

$ E[X^{1}]=\int_{-\infty }^{\infty }xf_{X}(x)dx=\frac{1}{3\sqrt{2\Pi }}\int_{-\infty }^{\infty }xe^{-\frac{(x-3)^{2}}{18}}dx=\frac{1}{3\sqrt{2\Pi }}9\sqrt{2\Pi }=3 $

Comment on Answer 1

I have seen 'E(x^n) is defined as:'

$ {\color{blue} E[X^{n}]=\int_{-\infty }^{\infty }x^{n}f_{X}(x)dx \;\;\; (1)} $

in a few of the hmwrk 5 answers. However, I think there's an important distinction between equivalency and definition. E(x^n) is not defined as (1); it is only E(x) that is defined as

$ {\color{blue} E[X]=\int_{-\infty }^{\infty }xf_{X}(x)dx \;\;\; (2)} $

E(x^n) happens to equal (1) by way of the more general fact that

$ {\color{blue} E[g(X)]=\int_{-\infty }^{\infty }g(X)f_{X}(x)dx \;\;\; (3)} $

See pgs 84-85 from Bertsekas and Tsitsiklis to see why. -ag

Instructor's comment: You are correct ag, I couldn't have said it better. I appreciate your attention to details. -pm

Another comment on Answer 1

Did you figure out the integral "by hand" or did you just plug it into a symbolic conputation software? You will need to be able to integrate by hand on the test. -pm

Answer 2

Write it here.

Answer 3

Write it here.

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