Conditional probability

  • #1

Homework Statement


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).


Homework Equations


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

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)?
 

Answers and Replies

  • #2
tiny-tim
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Hi BookMark440! :smile:

Hint: what is P(y>x) for a fixed value of x? :wink:
 
  • #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:
  • #4
tiny-tim
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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 ?
No, for P(y>x) you need to find the portion of y, don't you? :smile:
 

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