| Line 27: | Line 27: | ||
: <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> -TA | : <math>f_{X|A}(x|A)= \frac{f_{X}(x)}{P({X>3})} = \frac{f_{X}(x)}{1- F_{X}(3)} .</math> -TA | ||
| + | |||
| + | ===Comment on Hint=== | ||
| + | : <span style='color:blue'>It's important to note that the <math>\color{blue} f_X(x)</math> given in the final line of the hint is distinct from the pdf given in the problem statement. Specifically, the new <math>\color{blue} f_X(x)</math> is nonzero only on the range that the event A guarantees such that</span> | ||
| + | |||
| + | :<math>\color{blue} | ||
| + | f_X(x)=\left\{ | ||
| + | \begin{array}{ll} | ||
| + | c x^2, & 3<x<5,\\ | ||
| + | 0, & \text{ else}. | ||
| + | \end{array} | ||
| + | \right.</math> | ||
| + | |||
| + | :<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.</span> | ||
===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. | ||
Revision as of 17:52, 26 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!
Answer 1
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 that the event A guarantees 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.
Answer 2
Write it here.
Answer 3
Write it here.
