Exponential and Gamma Distribution: Component Failure Question

Question

Suppose that under normal operating conditions, the operating time until failure of a certain type of component has exponential($\lambda$) distribution for some $\lambda > 0$. And suppose that the random variables representing lifetimes of different components of this type may be regarded as independent.

Given the average lifetime of 10,000 components is found to be 20 days. Estimate the value of $\lambda$ based on this information

Textbook Answer:

Answer 1

Suppose $X_i \sim \mathsf{Exp}(\text{rate} = \lambda).$ Then $T = \sum_{i=1}^n X_i\sim \mathsf{Gamma}(\text{shape}=n,\, \text{rate}=\lambda)$ and $\bar X = \frac T n \sim \mathsf{Gamma}(\text{shape}=n,\, \text{rate}=n\lambda).$ Thus for $n$ observations, we have $E(X_i) = \frac{1}{\lambda},\,$ $E(T) = \frac{n}{\lambda},\,$ and $E(\bar X) = \frac{1}{\lambda}.$ [I'm using $T$ instead of $S$ because $S$ should be reserved notation for the sample standard deviation.]

However, $\hat \lambda = \frac n T $ is not an unbiased estimator of $\lambda.$ The following simulation suggests (does not prove) that $\check \lambda = \frac{n-1}{T}$ is an unbiased estimator of $\lambda,$ whereas $\hat \lambda = \frac n T$ is not. Formal proofs of the above relationships are via moment generating functions and (in the case of $E(\check \lambda)$) integration.

set.seed(318) # retain seed statement for exactly same simulation, omit for fresh run
m = 10^6;  n = 5;  lam = 1/10
t = replicate(m,  sum(rexp(n, lam)))
mean(t); mean(t/n); mean(n/t);  mean((n-1)/t)
## 49.99976   # aprx E(T) = 50
## 9.999951   # aprx E(avg) = 10
## 0.1250256  # n/T NOT unbiased for lambda 
## 0.1000204  # suggests E((n-1)/T) = lambda

However, for $n$ as large as 10,000, the bias of $\hat \lambda$ is negligible in most practical applications. So if $\bar X = T/n= 20,$ then $1/ \hat \lambda = n/T = 1/\bar X = 1/20$ is a reasonable (even if very slightly biased) estimate of $\lambda.$

The figure below illustrates that the sample mean $\bar X = \frac T n \sim \mathsf{Gamma}(\text{shape}=n,\, \text{rate}=n\lambda),$ as mentioned above; the gamma PDF is superimposed on a histogram of the simulated values of $\bar X.$