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Conditional PDF questions

  1. Oct 30, 2011 #1
    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. jcsd
  3. Oct 30, 2011 #2
    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.
     
  4. Oct 30, 2011 #3

    I like Serena

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    Welcome to PF, de1337ed! :smile:

    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: [itex]\int P(X=x|A) x dx = {\int P(X=x \& A) x dx \over P(A)}[/itex].

    For starters, can you calculate P(A)?
     
  5. Oct 30, 2011 #4
    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?
     
  6. Oct 30, 2011 #5
    You know that [itex]\int_{1}^{2} f_{X} \left(x\right)dx=1 [/itex] right?

    But [itex]\int_{1}^{2} 2x^{2}dx \neq 1 [/itex]. So what you need is some [itex] c [/itex] such that [itex]\int_{1}^{2}cx^{2}dx = 1[/itex]. Can you see how to do that?
     
  7. Oct 30, 2011 #6
    Oh shoot, copied the problem down wrong, its actually cx^-2
     
  8. Oct 30, 2011 #7
    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.
     
  9. Oct 30, 2011 #8

    I like Serena

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    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?
     
  10. Oct 30, 2011 #9
    So basically, I got that P(A) = 1/3.
    And fX|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?
     
  11. Oct 30, 2011 #10

    I like Serena

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    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.
     
  12. Oct 30, 2011 #11
    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
  13. Oct 30, 2011 #12

    I like Serena

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    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.
     
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