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= [[:Category:Problem solving|Practice Problem]]: Obtain the moment generating function for an exponential random variable = | = [[:Category:Problem solving|Practice Problem]]: Obtain the moment generating function for an exponential random variable = | ||
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| − | Let X be an exponential random variable. Recall that the pdf of an exponential random variable is given by | + | Let X be an exponential random variable. Recall that the pdf of an exponential random variable is given by |
| + | <br> <math>\ f_X(x)= \lambda e^{-\lambda x}, \text{ for }x\geq 0 . </math> | ||
| − | < | + | Obtain the moment generating function <span class="texhtml">''M''<sub>''X''</sub>(''s'')</span> of X. |
| − | |||
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=== Answer 1 === | === Answer 1 === | ||
| − | Hint: | + | |
| − | :<math>M_X(s)=E[e^{sX}] = \int e^{sx} f_{X}(x)dx</math> | + | Hint: |
| − | : get an answer for < | + | |
| − | : note that the integrating range of x starts from 0 | + | :<math>M_X(s)=E[e^{sX}] = \int e^{sx} f_{X}(x)dx</math> |
| + | :get an answer for <span class="texhtml">''s'' < λ</span> | ||
| + | :note that the integrating range of x starts from 0 | ||
| + | |||
=== Answer 2 === | === Answer 2 === | ||
| − | |||
| + | :<math>M_X(jw) = E[e^{jwx}]</math> | ||
| + | :<math>E[e^{jwx}] = \int e^{jwx} f_{X}(x)dx \text{ For } x \geq 0 </math> | ||
| + | :<math>M_X(jw) = \int_{0}^{\infty} e^{jwx} \lambda e^{-\lambda x}dx</math> | ||
| + | :<math>M_X(jw) = \lambda \int_{0}^{\infty} e^{jwx} e^{-\lambda x}dx</math> | ||
| + | :<math>M_X(jw) = \lambda \int_{0}^{\infty} e^{(jw-\lambda)x}dx</math> | ||
| + | :<math>M_X(jw) = \lambda \int_{0}^{\infty} e^{-(\lambda-jw)x}dx</math> | ||
| + | :<math>M_X(jw) = \lambda \frac {1}{-(\lambda-jw)}(e^{-(\lambda-jw)\infty}-e^{-(\lambda-jw)0})</math> | ||
| + | :<math>M_X(jw) = \lambda \frac {1}{-(\lambda-jw)}(e^{-\infty}-e^{0})</math> | ||
| + | :<math>M_X(jw) = \lambda \frac {1}{-(\lambda-jw)}(0-1)</math> | ||
| + | :<math>M_X(jw) = \frac {\lambda}{\lambda-jw}</math> | ||
| + | :<math>M_X(s) = \frac {\lambda}{\lambda-s}</math> | ||
| + | <br> | ||
| + | :<span style="color:green"> Instructor's comment: Be careful. In general, you can't obtain the Laplace transform by simply replacing <math>j\omega</math> by s. In fact, the corresponding integral for s diverges for some values of s. -pm </span> | ||
=== Answer 3 === | === Answer 3 === | ||
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Write it here. | Write it here. | ||
| + | |||
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[[2013 Spring ECE 302 Boutin|Back to ECE302 Spring 2013 Prof. Boutin]] | [[2013 Spring ECE 302 Boutin|Back to ECE302 Spring 2013 Prof. Boutin]] | ||
| − | '''[[ECE302|Back to ECE302]]''' | + | '''[[ECE302|Back to ECE302]]''' |
| + | |||
| + | [[Category:ECE302]] [[Category:ECE302Spring2013Boutin]] [[Category:Problem_solving]] [[Category:Probability]] [[Category:Discrete_random_variable]] [[Category:Characteristic_function]] | ||
Latest revision as of 13:00, 27 March 2013
Contents
Practice Problem: Obtain the moment generating function for an exponential random variable
Let X be an exponential random variable. Recall that the pdf of an exponential random variable is given by
$ \ f_X(x)= \lambda e^{-\lambda x}, \text{ for }x\geq 0 . $
Obtain the moment generating function MX(s) of X.
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
Hint:
- $ M_X(s)=E[e^{sX}] = \int e^{sx} f_{X}(x)dx $
- get an answer for s < λ
- note that the integrating range of x starts from 0
Answer 2
- $ M_X(jw) = E[e^{jwx}] $
- $ E[e^{jwx}] = \int e^{jwx} f_{X}(x)dx \text{ For } x \geq 0 $
- $ M_X(jw) = \int_{0}^{\infty} e^{jwx} \lambda e^{-\lambda x}dx $
- $ M_X(jw) = \lambda \int_{0}^{\infty} e^{jwx} e^{-\lambda x}dx $
- $ M_X(jw) = \lambda \int_{0}^{\infty} e^{(jw-\lambda)x}dx $
- $ M_X(jw) = \lambda \int_{0}^{\infty} e^{-(\lambda-jw)x}dx $
- $ M_X(jw) = \lambda \frac {1}{-(\lambda-jw)}(e^{-(\lambda-jw)\infty}-e^{-(\lambda-jw)0}) $
- $ M_X(jw) = \lambda \frac {1}{-(\lambda-jw)}(e^{-\infty}-e^{0}) $
- $ M_X(jw) = \lambda \frac {1}{-(\lambda-jw)}(0-1) $
- $ M_X(jw) = \frac {\lambda}{\lambda-jw} $
- $ M_X(s) = \frac {\lambda}{\lambda-s} $
- Instructor's comment: Be careful. In general, you can't obtain the Laplace transform by simply replacing $ j\omega $ by s. In fact, the corresponding integral for s diverges for some values of s. -pm
Answer 3
Write it here.
