Finding the MSE of some Estimator $\hat{\theta}$ the best way possible.

Question

I'd glad if you could help me. I was given the following question in my exam, but i just couldn't finish that on time. I wonder if there's some easier way to solve this.

Given the following random samples $(X_1, X_2,...,X_n)$ of independent random variables with the same probability density function: $$f_X(x)= \begin{cases} e^{\lambda - x}, & \text{$x\ge \lambda$ } \\ 0, & \text{$x < \lambda$} \end{cases}$$ where $\lambda >0$.

We want to estimate the unknown parameter $\theta=\lambda$.

We are given with the following Estimator: $\hat{\theta}=X_{(1)}-\frac{1}{n}$

where $X_{(1)}=min\{X_1,X_2,...,X_n\}$

and we need to find it's MSE (Mean Square Error: $E\big((\hat{\theta}-\lambda)^2\big)$.

I already solve this problem, but that was too complicated. I wonder if there's a "catch". I don't know, maybe we can identify some familiar distributaion along the way, something that can help us find the Expected Value much quicker. That's usually the case.

My calculation was:

1. Fiding CDF of $X_{(1)}: \quad F_{X_{(1)}}(t)=1-e^{n(\lambda -x)}$

2. Fiding PDF of $X_{(1)}: \quad f_{X_{(1)}}(t)=ne^{n(\lambda -x)}$

3. Fiding Expected Value of $X_{(1)}:$ $$ E(X_{(1)})= \int_{\lambda}^{\infty}xne^{n(\lambda -x)}dx= \text{...long calculation...} =\lambda + \frac{1}{n}$$ here we can see that $\hat{\theta}=X_{(1)}-\frac{1}{n}$ is unbiased: $E\big(\hat{\theta} \big) = E \big( X_{(1)}-\frac{1}{n} \big) = E\big( X_{(1)}\big) - E\big( \frac{1}{n} \big) = \lambda + \frac{1}{n} - \frac{1}{n} = \lambda = \theta$, so one last thing to do is to find the Variance of $\hat{\theta}$:

4. Find $$ V(\hat{\theta}) = V( X_{(1)}-\frac{1}{n} ) = V( X_{(1)} ) = E \big( (X_{(1)} - \lambda)^2 \big) = \int_{\lambda}^{\infty} {(x-\lambda)^2ne^{n(\lambda -x)}}dx= \text{...too long calculation. I calculated using WolframAlpha} = \frac{2}{n^2} = MSE(\hat{\theta})$$

Answer 1

Perhaps a quicker way:

$X_i-\lambda$ has an exponential distribution with rate $1$

so $X_{(1)}-\lambda$ has an exponential distribution with rate $n$

so $\mathbb E[X_{(1)}-\lambda]=\frac1n$ and $Var(X_{(1)}-\lambda)=\frac1{n^2}$

and $Var(X_{(1)}-\lambda)= \mathbb E\left[\left(X_{(1)}-\lambda -\frac1n\right)^2\right] = \mathbb E\left[\left(\left(X_{(1)} -\frac1n\right) -\lambda\right)^2\right]$,

which is the mean square error, and we already know this variance is $\frac1{n^2}$