Thread: Probability and statistics

The time X to the first click on Geiger counter has exponential distribution f(x)=e^-x , x>0. The time Y from the first click to the second has the same exponential distribution as X and is independent of X.
1. write down the joint p.d.f. of X and Y and show that, for t>0,
P(Y<tX) = t/(1+t) and
2. deduce that, for 0<u<1; P(X/(X+Y)<u)=u
3. interpret the result in 2 above...

Hey samuel2012.

Can you show us what you have tried?

Hint: If X and Y are independent then P(X = x AND Y = y) = P(X = x)*P(Y = y). where P(X = x) = e^(-x) and similarly for P(Y = y).

Now you need to consider the region for Y < tX. This means you will need to look at the region of integration where this happens. Since t > 0 is given and since this is a linear relationship, if you use a double integral then you can find the region under the triangle.

If X = 0 then Y = 0. If X is infinity then Y is infinity as well. Now you will get a line from 0 to infinity with a gradient equal to t and the region will be between the x-axis and this line.

Using this hint, can you construct an integral with the right integration region to get your answer?

1. f(x,y) = e^-(x+y)

P(Y<tx|X=x).P(X=x) = (1-e^-tx)(e^-x)


P(Y<tx) = ∫(1-e^-tx)(e^-x)dx
= ∫[e^-x - e^-x(t+1)]dx

= 1-1/(t+1)
=t/(1+t)


2. P(X(1-u)<uY)=u

P(X(1-u)<uY) = u
P(Y>X(1-u)/u) = u

P(Y<tX) = 1-u where t = (1-u)/u