I have two questions.

1) If $X_1,X_2,X_3,...,X_n$ constitute a random sample of size $n$ from an exponential distribution, show that $\bar X$ is a consistent estimator of the parameter $\lambda$.

This is my attempt:
Mean of exponential distribution is $\lambda^{-1}$. Thus by central Limit theorem $E(\bar X)=\lambda^{-1}$. Using mean square consistency, if $\lim_{n\to \infty} E[(\bar X -\lambda )^2]=0$ then $\bar X $ is consistent for $\lambda$. Since $$ E[(\bar X -\lambda )^2]= E[(\bar X)^2]+E(\lambda)^2-2\lambda E(\bar X) , $$ I get $$ \lim_{n\to \infty} E[(\bar X -\lambda )^2] = \lim_{n\to \infty} {2 \over \lambda ^2 }+\lambda^2-2 . $$ Clearly this is not equal to $0$. How can I show this is consistent? I tried using Chebechev inequality as well. But even using that I couldn't.

2) Is $X_{(n)}$ a consistent estimator for $\lambda$?
If $P[|X_{(n)}-\lambda|<\epsilon]$ approaches $1$ as $n$ goes to infinity for all $\epsilon > 0$ then $X_n$ is a consistent estimator for $\lambda$.

The probability density function of $X_{(n)}$ is $f_n(x) = n \lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}$, for $X>0$.

Thus \begin{align*} P[|X_{(n)}-\lambda|<\epsilon] &= P[\lambda -\epsilon<X_{n}<\lambda+\epsilon] \\ &=P[0<X_{n}<\lambda+\epsilon] \\ &=\int_0^{\lambda+\epsilon} n \lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}\\ &= (1-e^{- \lambda X_{(n)} }), \end{align*} where $X{(n)}$ takes values in $[0 , \lambda+ \epsilon]$.

Thus, can I say $\lim_{n\to \infty}[1-e^{-\lambda(\lambda+\epsilon)}]^n=1$?


1 Answer 1


Question 1: $\bar X$ cannot be a consistent estimator of $\lambda$ but since $$\text{plim} \bar X = \lambda ^{-1} \Rightarrow \frac 1 {\text{plim} \bar X} = \lambda$$ we can recover the true $\lambda$ by considering the reciprocal of the probability limit of $\bar X$. (except if the original question uses the inverse parametrization for the Exponntial distribution, where we specify a scale rather than a rate parameter -but then again, the $\lambda$-notation is usually employed for the rate parameter...)

Question 2: Intuitively, the maximum order statistic cannot be a consistent estimator, since $\lambda$ is finite while the support of the Exponential distribution is open from above, so the maximum order statistic will tend to infinity as the sample size tends to infinity. Using the formal approach, the full result is

$$P[|X_{(n)}-\lambda|<\epsilon] = (1-e^{- \lambda (\lambda+ \epsilon) })^n - (1-e^{- \lambda (\lambda- \epsilon) })^n\cdot I_{\{\lambda>\epsilon\}}$$


$$P[|X_{(n)}-\lambda|<\epsilon] =\begin{cases} (1-e^{- \lambda (\lambda+ \epsilon) })^n - (1-e^{- \lambda (\lambda- \epsilon) })^n, & \lambda>\epsilon \\ (1-e^{- \lambda (\lambda+ \epsilon) })^n, &\lambda \leq\epsilon \end{cases}$$

In all cases the involved exponential expressions are positive and smaller than unity, therefore "one minus them in the $n$-th power" goes to zero, not to unity as would be required for consistency. In other words the distance between the maximum order statistic and any finite $\lambda$ grows unbounded with probability one.


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