| Line 26: | Line 26: | ||
: Find c by, | : Find c by, | ||
: <math>\int_{-\infty}^{\infty} f_{X}(x)dx =1.</math> | : <math>\int_{-\infty}^{\infty} f_{X}(x)dx =1.</math> | ||
| − | : <math>f_{X|A}(x|A)= \frac{f_{X}(x)}{P({X>3})} = \frac{f_{X}(x)}{1- F_{X}(3)} | + | : <math>f_{X|A}(x|A)= \frac{f_{X}(x)}{P({X>3})} = \frac{f_{X}(x)}{1- F_{X}(3)}</math> for some range of x |
| + | -TA | ||
===Comment on Hint=== | ===Comment on Hint=== | ||
Latest revision as of 08:17, 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)} $ for some range of x
-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.
