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Conditional probability

  1. Jul 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Given f(x) = e^-x and f(y|x) = 1/x e^(-y/x). Three parts: (a) Compute density of (x,y), (b) Compute E(y) and (c) Compute P(y>x).


    2. Relevant equations
    f(x,y) = f(y|x)f(x)
    if f(x) = ve^(-vx), then E(x)=v^(-1)

    3. The attempt at a solution

    I'm stuck on a problem. I was given f(x) and f(y|x) and was able to derive f(x,y) and compute E(y). The third step of the problem is computing P[y>x]. I think I need to know f(y) to answer this problem but I can't figure out how to derive it. Or is there a way to compute P(y>x) given the info I know without deriving f(y)?
     
  2. jcsd
  3. Jul 30, 2009 #2

    tiny-tim

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    Hi BookMark440! :smile:

    Hint: what is P(y>x) for a fixed value of x? :wink:
     
  4. Jul 30, 2009 #3
    Maybe I am looking for the wrong answer. Do I simply need to find the portion of f(x)=exp(-x) that are to the left of its intersection with f(x)=x ?

    This would mean I am looking for the intersection, which is exp(-x) = x, solving for x?

    That is:

    -x*log e = log x
    0.4343 = -(log x)/x

    But what step is next? Is there some law of logs I am missing to simplify this? From graphing the problem, it looks like the point of intersection is (1 - log e) but I can't see how the above translate into (1-log e).
     
    Last edited: Jul 31, 2009
  5. Jul 31, 2009 #4

    tiny-tim

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    No, for P(y>x) you need to find the portion of y, don't you? :smile:
     
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