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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!
 
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!
 
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===Answer 1===
+
===Hint===
 
Hint:
 
Hint:
 
: Find c by,
 
: Find c by,
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:<span style='color:blue'>Note that this is 'new' <math>\color{blue} f_X(x)</math> is not a valid pdf by itself (violates normalization to 1 axiom), and thus the normalizing denominator is used. -ag</span>
 
:<span style='color:blue'>Note that this is 'new' <math>\color{blue} f_X(x)</math> is not a valid pdf by itself (violates normalization to 1 axiom), and thus the normalizing denominator is used. -ag</span>
===Answer 2===
+
 
 +
**<span style='color:green'>Correct. -pm</span>
 +
 
 +
===Answer 1===
 
Write it here.
 
Write it here.
===Answer 3===
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===Answer 2===
 
Write it here.
 
Write it here.
 
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Revision as of 03:52, 27 March 2013

Practice Problem: What is the conditional density function


Let X be a continuous random variable with probability density function

$ f_X(x)=\left\{ \begin{array}{ll} c x^2, & 1<x<5,\\ 0, & \text{ else}. \end{array} \right. $

Let A be the event $ \{ X>3 \} $. Find the conditional probability density function $ f_{X|A}(x|A). $


Share your answers below

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!


Hint

Hint:

Find c by,
$ \int_{-\infty}^{\infty} f_{X}(x)dx =1. $
$ f_{X|A}(x|A)= \frac{f_{X}(x)}{P({X>3})} = \frac{f_{X}(x)}{1- F_{X}(3)} . $ -TA

Comment on Hint

It's important to note that the $ \color{blue} f_X(x) $ given in the final line of the hint is distinct from the pdf given in the problem statement. Specifically, the new $ \color{blue} f_X(x) $ is nonzero only on the range dictated by the occurrence of event 'A' such that
$ \color{blue} f_X(x)=\left\{ \begin{array}{ll} c x^2, & 3<x<5,\\ 0, & \text{ else}. \end{array} \right. $
Note that this is 'new' $ \color{blue} f_X(x) $ is not a valid pdf by itself (violates normalization to 1 axiom), and thus the normalizing denominator is used. -ag
    • Correct. -pm

Answer 1

Write it here.

Answer 2

Write it here.


Back to ECE302 Spring 2013 Prof. Boutin

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