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===Answer 1=== | ===Answer 1=== | ||
| − | + | The moment of order n is defined as | |
| + | <math>E(X^n)=\int_{-\infty}^{\infty} x^n*f_X(x)</math> | ||
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| + | since <math> x^0 = 1 </math> and <math>\int_{-\infty}^{\infty} f_X(x) = 1 </math> | ||
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| + | the moment of order zero is <math>E \left( X^0 \right) = 1</math> | ||
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| + | :<span style="color:purple"> Instructor's comment: Don't forget to put the "dx" in the integral. Also, I should warn you that the symbol "*" denotes convolution. I believe you mean "multiplication", right? Can anobody write a more "compact" solution? -pm </span> | ||
===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. | ||
Latest revision as of 03:47, 4 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 zero of that random variable. In other words, compute
$ E \left( X^0 \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 order n is defined as $ E(X^n)=\int_{-\infty}^{\infty} x^n*f_X(x) $
since $ x^0 = 1 $ and $ \int_{-\infty}^{\infty} f_X(x) = 1 $
the moment of order zero is $ E \left( X^0 \right) = 1 $
- Instructor's comment: Don't forget to put the "dx" in the integral. Also, I should warn you that the symbol "*" denotes convolution. I believe you mean "multiplication", right? Can anobody write a more "compact" solution? -pm
Answer 2
Write it here.
Answer 3
Write it here.
