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= [[:Category:Problem solving|Practice Problem]]: obtaining the joint pdf from the marginals of two independent variables = | = [[:Category:Problem solving|Practice Problem]]: obtaining the joint pdf from the marginals of two independent variables = | ||
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Another random variable Y has the following probability density function: | Another random variable Y has the following probability density function: | ||
| − | <math> f_Y (y) = \frac{1}{3 \sqrt{2\pi} } e^{\frac{-(x-7)^2}{6}}.</math> | + | <math> f_Y (y) = \frac{1}{3 \sqrt{2\pi} } e^{\frac{-(x-7)^2}{6}}.</math> |
| − | Assuming that X and Y are independent, find the joint probability function <span class="texhtml">''f''<sub>'' | + | Assuming that X and Y are independent, find the joint probability function <span class="texhtml">''f''<sub>''XY'''''</sub>'''''('''''<b>x'',''y'').'' |
| + | </b> | ||
| + | <br> | ||
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| + | *<span style="color:red">- Is this a valid pdf (coeff divided by 3 instead of sqrt(3)), or is it unnecessary to satisfy prob. axioms here?</span> | ||
| + | **<span style="color:purple">Good point. That's a typo. I'm glad you pointed it out. Can somebody please find the correct normalization? -pm </span> | ||
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Because X and Y are independent, the joint probability function can be represented as the product of the two marginal density functions:<br> | Because X and Y are independent, the joint probability function can be represented as the product of the two marginal density functions:<br> | ||
| − | < | + | <math>f_{XY}(x,y) = f_X(x)f_Y(y)</math> |
Thus, the joint probability function is simply the two marginal density functions multiplied together: | Thus, the joint probability function is simply the two marginal density functions multiplied together: | ||
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<math>f_{XY}(x,y) = \frac{1}{6\pi} e^{\frac{1}{6}(-4x^2+14x-49)}.</math> | <math>f_{XY}(x,y) = \frac{1}{6\pi} e^{\frac{1}{6}(-4x^2+14x-49)}.</math> | ||
| + | :<span style="color:purple"> Instructor's comment: Do you know why the joint pdf is the product of the marginals? All we have seen in class is that the probability of two independent events is the product of the probabilities of the respective events. As you know, the pdf is not a probability. -pm </span> | ||
| + | |||
| + | :<span style="color:blue"> Is this because the pdf is the envelope function of the probabilities just as the Fourier transform X(jw) is the envelope function of the set of Fourier coefficients {a_k}, where <math>prob(x) = \int_{x^-}^{x^+}f_X(x) dx</math>? -ag </span> | ||
| + | ::<span style="color:purple"> Sorry, I am not sure what you mean by "envelop function." -pm </span> | ||
=== Answer 2 === | === Answer 2 === | ||
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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:Continuous_random_variable]] | [[Category:ECE302]] [[Category:ECE302Spring2013Boutin]] [[Category:Problem_solving]] [[Category:Continuous_random_variable]] | ||
Latest revision as of 12:03, 26 March 2013
Contents
Practice Problem: obtaining the joint pdf from the marginals of two independent variables
A random variable X has the following probability density function:
$ f_X (x) = \frac{1}{\sqrt{2\pi}} e^{\frac{-x^2}{2}}. $
Another random variable Y has the following probability density function:
$ f_Y (y) = \frac{1}{3 \sqrt{2\pi} } e^{\frac{-(x-7)^2}{6}}. $
Assuming that X and Y are independent, find the joint probability function fXY(x,y).
- - Is this a valid pdf (coeff divided by 3 instead of sqrt(3)), or is it unnecessary to satisfy prob. axioms here?
- Good point. That's a typo. I'm glad you pointed it out. Can somebody please find the correct normalization? -pm
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
Because X and Y are independent, the joint probability function can be represented as the product of the two marginal density functions:
$ f_{XY}(x,y) = f_X(x)f_Y(y) $
Thus, the joint probability function is simply the two marginal density functions multiplied together:
$ f_{XY}(x,y) = \frac{1}{6\pi} e^{\frac{1}{6}(-4x^2+14x-49)}. $
- Instructor's comment: Do you know why the joint pdf is the product of the marginals? All we have seen in class is that the probability of two independent events is the product of the probabilities of the respective events. As you know, the pdf is not a probability. -pm
- Is this because the pdf is the envelope function of the probabilities just as the Fourier transform X(jw) is the envelope function of the set of Fourier coefficients {a_k}, where $ prob(x) = \int_{x^-}^{x^+}f_X(x) dx $? -ag
- Sorry, I am not sure what you mean by "envelop function." -pm
Answer 2
Write it here.
Answer 3
Write it here.
