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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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− | === | + | ===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 | + | |
+ | **<span style='color:green'>Correct. -pm</span> | ||
+ | |||
+ | ===Answer 1=== | ||
Write it here. | Write it here. | ||
− | ===Answer | + | ===Answer 2=== |
Write it here. | Write it here. | ||
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Revision as of 03:52, 27 March 2013
Contents
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). $
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.