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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'').''</span>  
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Assuming that X and Y are independent, find the joint probability function <span class="texhtml">''f''<sub>''X''''Y'''''</sub>'''''('''''<b>x'',''y'').''&lt;/span&gt;
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</b></span>
  
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'''
 
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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>  
  
<span class="texhtml">''f''<sub>''X''''Y'''</sub>'''(''x'',''y'') = ''f''<sub>''X''</sub>(''x'')''f''<sub>''Y''</sub>(''y'')'''</span>  
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<span class="texhtml"</span><math>f_{XY}(x,y) = f_X(x)f_Y(y)</math><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:  
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[[2013 Spring ECE 302 Boutin|Back to ECE302 Spring 2013 Prof. Boutin]]  
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[[2013 Spring ECE 302 Boutin|Back to ECE302 Spring 2013 Prof. Boutin]]
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'''
  
[[ECE302|Back to ECE302]]  
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'''[[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]]

Revision as of 17:26, 1 March 2013

[[Category:independent random variables

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>


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Answer 1

Because X and Y are independent, the joint probability function can be represented as the product of the two marginal density functions:

<span class="texhtml"</span>$ 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)}. $

Answer 2

Write it here.

Answer 3

Write it here.


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