# Conditional PDF questions

1. Oct 30, 2011

### de1337ed

The random variable X has the PDF
fX(x) = cx^-2 if 1 < x < 2;
0 otherwise:
(a) Determine the value c
(b) Let A be the event fX > 1.5. Calculate the conditional expectation and the
conditional variance of X given A.

I'm not understanding the concept of conditional pdf's. Can someone show me the step by step?
Also what would be the ranges for the conditional expectation. I know that fX(x) is from 1<x<2, and E[x] is taken from >1.5, so its range 0<x<0.5?
Also, because we want to calculate the even fX > 1.5, does that involve using the CDF?
Thank you.

Last edited: Oct 30, 2011
2. Oct 30, 2011

### Cant or Wont

Someone posting the step by step will do little to help your understanding. How far have you got?

Also this probably belongs in the Homework and Coursework forum and not here.

3. Oct 30, 2011

### I like Serena

Welcome to PF, de1337ed!

Did you already find the value for c?

As for conditional pdf's, do you know the formula P(B|A)=P(A & B) / P(A)?
And the formula for expectation E[X] = ∫P(x) x dx?

The conditional expectation given A would be: $\int P(X=x|A) x dx = {\int P(X=x \& A) x dx \over P(A)}$.

For starters, can you calculate P(A)?

4. Oct 30, 2011

### de1337ed

Ya I found the value of c to be 2. And ya, I know that is the formula, but I didn't know how to calculate P(A)... Is it simply the ∫fX(x)dx from 1.5 to 2?

5. Oct 30, 2011

### Cant or Wont

You know that $\int_{1}^{2} f_{X} \left(x\right)dx=1$ right?

But $\int_{1}^{2} 2x^{2}dx \neq 1$. So what you need is some $c$ such that $\int_{1}^{2}cx^{2}dx = 1$. Can you see how to do that?

6. Oct 30, 2011

### de1337ed

Oh shoot, copied the problem down wrong, its actually cx^-2

7. Oct 30, 2011

### Cant or Wont

Sorry my bad you actually copied it correctly! It's been a long day! 2 is correct then. And yes that is the correct way to get the probability of A.

8. Oct 30, 2011

### I like Serena

Good! :)

So what is P(A)?

And what would the range of x be for which P(X=x & A) has a non-zero value?

9. Oct 30, 2011

### de1337ed

So basically, I got that P(A) = 1/3.
and 0 otherwise

Then E[x] = ∫x*6x-2dx. From 1.5 to 2.
And then Variance would be E[x2] - (E[x])2

Am I right?

10. Oct 30, 2011

### I like Serena

Except for the range you are right.

The range for fX is 1<x<2.
With A given, you get that x>1.5, so the relevant range of fX|A is 1.5<x<2.

This is what you used in your formula for E[x], so I don't get why you'd think it is 0<x<0.5.

11. Oct 30, 2011

### de1337ed

Hmm, i see, okay. I think I'm doing something wrong,

I'm getting the fact that E[x] = 1.726...
E[x2] = ∫x2*2x-2dx from 1.5 to 2. This gives me 1

As a result, I get the variance to be 1-(1.726)2 = -1.979. Which doesn't make sense because isn't the variance supposed to be ≥0?

EDIIITTT::: NVM, i might be an idiot. Haven't slept, I'm using the wrong PDF, lol

So what I ended up doing was var(x) = ∫(x-1.726)2*6x-2dx = 0.0205

Seem good?

Last edited: Oct 30, 2011
12. Oct 30, 2011

### I like Serena

E[X|A] is good. :)

For you variance you have used the original pdf instead of the pdf of X|A.

And yes, the variance is supposed to be at least zero.