[Math] Exponential Distribution ( Probability Problem ).

Question

We know that probability density function $f(x)$ for an exponential distribution with parameter $\lambda$ is given by :

$f(x)= \lambda e^{- \lambda x}$

We are given the following question :

If the number of minutes it takes for a mechanic to check a tyre is a random variable having an exponential distribution with mean 5 , what is the probability that the mechanic will take more than 8 minutes to check 2 tyres ?

My take to the problem :

Mean time ( 1 Tyre ) = 5 ,

Mean time (2 Tyres ) = 10.

So , $\frac{1}{\lambda}=10$ , as for an exponential distribution Mean = $\frac{1}{\lambda}$.

With the above information , I integrate $f(x)$ with limits from $8$ to $\infty$ ,
But I am not getting the correct answer..

Could anyone tell , what am I doing wrong ?

Answer 1

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