Conditional PDF and Expectation: How to Calculate Them?

[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.

 

[Cant or Wont]

I'm not understanding the concept of conditional pdf's. Can someone show me the step by step?

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.

 

[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)?

 

[de1337ed]

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)?

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?

 

[Cant or Wont]

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?

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?

 

[de1337ed]

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

 

[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.

 

[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?

 

[de1337ed]

So basically, I got that P(A) = 1/3.
And f_X|A(x|a) = 6x^-2, 0<x<0.5. (This range confuses me, I'm not sure about this)
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?

 

[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.

 

[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?

 

[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.