exponential distributionprobabilityprobability distributions
Two independent events A and B follow exponential distribution with parameter L:
f(t)=Le^(-Lt) for t>=0. If X is the time where A occurs and Y the time where B occurs, calculate the probability P[X>=2*Y] meaning that A happens at least after double the time that B occurred.
How do I proceed?
I need to integrate f(t) in the area where x>=2*y. Does that mean that 0<=y<=x/2?
It is easy to use moment generating functions to find the sum of two exponential distributions with the same rate $\lambda$ is $Gamma(shape=2, rate=\lambda)$. Of course, for that you need to know the MGFs.
The MGF of an exponential random variable is $m_X(t) = \frac{\lambda}{\lambda - t},$ for $ t < \lambda.$ For independent exponentials with the same rate the MGF for $Y = X_1 + X_2$ is $$m_Y(t) = [m_X(t)]^2 = \left(\frac{\lambda}{\lambda - t}\right)^2.$$ This is the MGF of $Gamma(shape=2, rate=\lambda).$
You can check density functions, MGFs, and so on in Wikipedia articles on 'exponential distribution' and 'gamma distribution', but be aware that the exponential distribution can be parameterized using the mean and the rate, and that the gamma distribution also has an alternate parameterization.
In a statistical computer package such as R it is easy to find
the answer to your question. The R code 1 - pgamma(8, 2, 1/5)
returns 0.5249309. Otherwise, you have some integration to do.
Below is a brief simulation that performs the two tire experiment a million times, giving results that are accurate to a few significant digits. Histograms show simulated values, and curves show densities.
m = 10^6; x1 = rexp(m, 1/5); x2 = rexp(m, 1/5); t = x1+x2
mean(x1); sd(x1); mean(x2)
## 5.007576 # approx E(X_1)
## 5.00329 # approx SD(X_1)
## 5.001606 # approx E(X_2)
mean(t); sd(t); mean(t > 8)
## 10.00918
## 7.07116
## 0.525946
The Poisson process can be interpreted as the process that counts the total number of events that have occurred when the waiting times between events are i.i.d. exponential.
Hence, if you find the holding/waiting times between some random events are i.i.d. and distributed exponentially with parameter $\lambda$, then the total number of events will be distributed as a Poisson distribution with parameter $\lambda t$.
Best Answer
Assuming $X$ and $Y$ to be random variables, you need to integrate the joint distribution of $(X,Y)$ which is easy to compute as they are independent. Your integral limits are fine: $0\leq x<\infty, 0\leq y\leq x/2$. I think the final answer is $1/3$.