Conditional probability and sum of rvs question

  • Thread starter MrFancy
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  • #1
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I'm trying to solve a problem as part of my research and it's giving me fits. It seems like it should be simple, but I can't wrap my brain around how to do it. The problem is:

Suppose X~N(0,s), and Y is a random variable that has a probability mass point at 0 but is otherwise uniformally distributed on (0,t] so that:

f(y)=k, y=0
f(y)=(1-k)(1/t), 0 < y < t
f(y)=0 otherwise

What is

Pr(y < A | x + y > B)

where A and B are arbitrary constants?

I think I've calculated the convolution of X and Y, but I'm not sure how to get the density from there (and I'm not sure I have the convolution right either). Thanks for any help you can provide.
 

Answers and Replies

  • #2
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Does this look like something that isn't going to have an analytical solution and would need to be simulated?
 
  • #3
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Try using the law of total probability to get

[tex] Pr(Y<a|X+Y>b)=\int_{-\infty}^{\infty} Pr(Y<a|Y>b-x) Pr(X=x) dx [/tex]

Sorry for the sloppy notation, but if you change the limits of the integral, I think you should be able to compute the answer.
 
  • #4
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OK, thanks a lot, I hadn't thought of using the law of total probability.

Here's what I did, simplifying a little so that Y has a mass point at 0 and is otherwise U[0,1] instead of U[0,t]:

For the Pr(Y<a|Y>b-x) part,
1. take the cdf of Y evaluated at a, which is a(1-k)+k
2. subtract the cdf of Y evaluated at b-x, which is (b-x)(1-k)+k
3. divide by (1- the cdf of Y evaluated at b-x), which is 1-((b-x)(1-k)+k)

The bounds of integration should be b-a to positive infinity, since this probability is 0 if x<(b-a). Then the Pr(X=x) part is just the pdf of a normal random variable. This gives me:

[tex] Pr(Y<a|X+Y>b)=\int_{b-a}^{\infty} (1-k)\frac{(a-b+x)}{1-(b-x)(1-k)-k}\frac{1}{\sigma\sqrt{2\pi}}exp(-\frac{e^2}{2\sigma^2}) dx [/tex]

Does that look right? If so, isn't that integral too messy to evaluate?
 
  • #5
283
3
OK, thanks a lot, I hadn't thought of using the law of total probability.

Here's what I did, simplifying a little so that Y has a mass point at 0 and is otherwise U[0,1] instead of U[0,t]:

For the Pr(Y<a|Y>b-x) part,
1. take the cdf of Y evaluated at a, which is a(1-k)+k
2. subtract the cdf of Y evaluated at b-x, which is (b-x)(1-k)+k
3. divide by (1- the cdf of Y evaluated at b-x), which is 1-((b-x)(1-k)+k)

The bounds of integration should be b-a to positive infinity, since this probability is 0 if x<(b-a). Then the Pr(X=x) part is just the pdf of a normal random variable. This gives me:

[tex] Pr(Y<a|X+Y>b)=\int_{b-a}^{\infty} (1-k)\frac{(a-b+x)}{1-(b-x)(1-k)-k}\frac{1}{\sigma\sqrt{2\pi}}exp(-\frac{e^2}{2\sigma^2}) dx [/tex]

Does that look right? If so, isn't that integral too messy to evaluate?
What you did looks correct except that it should be exp (x...) but i think that was a typo. I don't know if you can integrate that or not, certainly you can approximate it using numerical methods. If you want to solve the integral I suggest using a software or asking someone else as I suck at calculus and woke up quite early this morning. Good luck!
 

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