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<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>''X''''Y'''''</sub>'''''('''''<b>x'',''y'').'' | + | Assuming that X and Y are independent, find the joint probability function <span class="texhtml">''f''<sub>''X''''Y'''''</sub>'''''('''''<b>x'',''y'').'' |
| − | </b | + | </b> |
<br> | <br> | ||
Revision as of 17:34, 1 March 2013
[[Category:independent random variables
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 fX'Y(x,y).
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:
fX'Y(x,y) = fX(x)fY(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)}. $
Answer 2
Write it here.
Answer 3
Write it here.
