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<math>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</math> | <math>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</math> | ||
| − | <span style="color: | + | <span style="color:blue">I have seen 'E(x^n) is defined as:'</span> |
<math>E[X^{n}]=\int_{-\infty }^{\infty }x^{n}f_{X}(x)dx \;\;\; (1)</math> | <math>E[X^{n}]=\int_{-\infty }^{\infty }x^{n}f_{X}(x)dx \;\;\; (1)</math> | ||
| − | <span style="color: | + | <span style="color:blue">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</span> |
<math>E[X]=\int_{-\infty }^{\infty }xf_{X}(x)dx \;\;\; (2)</math> | <math>E[X]=\int_{-\infty }^{\infty }xf_{X}(x)dx \;\;\; (2)</math> | ||
| − | <span style="color: | + | <span style="color:blue">E(x^n) happens to equal (1) by way of the more general fact that</span> |
<math>E[g(X)]=\int_{-\infty }^{\infty }g(X)f_{X}(x)dx \;\;\; (3)</math> | <math>E[g(X)]=\int_{-\infty }^{\infty }g(X)f_{X}(x)dx \;\;\; (3)</math> | ||
| − | <span style="color: | + | <span style="color:blue">See pgs 84-85 from Bertsekas and Tsitsiklis to see why. -ag</span> |
=== Answer 2 === | === Answer 2 === | ||
Revision as of 21:28, 24 March 2013
Contents
Practice Problem: compute the zero-th 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) . $
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
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 $
I have seen 'E(x^n) is defined as:'
$ 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
$ 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
$ 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
Answer 2
Write it here.
Answer 3
Write it here.
