Line 16: | Line 16: | ||
</b></span> | </b></span> | ||
− | + | <br> | |
---- | ---- | ||
Line 30: | Line 30: | ||
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> | ||
− | <span class="texhtml">''f''<sub>''X''''Y''</sub>(''x'',''y'') = ''f''<sub>''X''</sub>(''x'')''f''<sub>''Y''</sub>(''y'')</span><br> | + | <span class="texhtml">''f''<sub>''X''''Y'''</sub>'''(''x'',''y'') = ''f''<sub>''X''</sub>(''x'')''f''<sub>''Y''</sub>(''y'')'''</span>'''<br> ''' |
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: |
Revision as of 17:31, 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).</span>
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.