UMVUE of $P(X_1 \ge t)$ for a two-parameter exponential distribution

Question

I'm attempting to find

$(a)$ The UMVUE of $\lambda$ when $\theta$ is known.

$(b)$ The UMVUE of $\theta$ when $\lambda$ is known.

$(c)$ The UMVUE of $P(X_1 \ge t)$ for a fixed $t > \theta$ when $\lambda$ is known.

I'm new to the concept of UMVUE and attempting to self-learn it through a mathematical statistics textbook. I would appreciate some feedback for $(a)$ and $(b)$ in terms of their correctness and some help with $(c)$.

I found a sufficient statistic $T = (X_{(1)}, \sum\limits_{i = 1 }^n {X_i })$

For $(a)$, when $θ$ is known, $\sum\limits_{i = 1 }^n {X_i }$ is a sufficient and complete statistic for $λ$.

$E(\sum\limits_{i = 1 }^n {X_i }) = n(\lambda + \theta)$

Therefore $T_1 = \frac{\sum\limits_{i = 1 }^n {X_i }}{n} - \theta = \bar X - \theta$ is the UMVUE of $\lambda$.

For $(b)$, when $\lambda$ is known, $X_{(1)}$ is sufficient and complete for $\theta$.

$E(X_{(1)}) = \lambda + \theta$. Therefore $T_2 = X_{(1)} - \lambda$ is the UMVUE of $\theta$.

For $(c)$, I'm not completely sure how to go about doing this but I'm assuming that the UMVUE would be $P(X_1 \ge t\mid T)$ and that it would be 1 when $t<X_{(1)}$ but I'm unsure how to deal with the other case and whether this is indeed correct.

Answer 1

Your answer for (a) is correct.

Now note the pdf of $X_{(1)}$:

$$f_{X_{(1)}}(z)=\frac{n}{\lambda}e^{-n(z-\theta)/\lambda}\mathbf1_{z>\theta}\quad, \,\theta\in\mathbb R$$

So for (b), $X_{(1)}-\frac{\lambda}{\color\red{n}}$ is the UMVUE of $\theta$.

In (c), $\lambda$ is known so your complete sufficient statistic is $X_{(1)}$.

By Lehmann-Scheffe, an unbiased estimator of $P(X_1\ge t)$ which is a function of $X_{(1)}$ will be the UMVUE of $P(X_1\ge t)$. Let $g(\cdot)$ be that function.

For a fixed $t>\theta$, $$P(X_1\ge t)=e^{-(t-\theta)/\lambda}$$

We have $$E\left[g(X_{(1)})\right]=P(X_1\ge t)\quad,\,\forall\,\theta$$

That is, for any $t>\theta$,

$$\frac{n}{\lambda}\int_{\theta}^\infty g(z)e^{-\frac{n}{\lambda}(z-\theta)}\,dz=e^{-(t-\theta)/\lambda}\quad,\,\forall\,\theta$$

Or, $$\int_{\theta}^\infty g(z)e^{-nz/\lambda}\,dz=\frac{\lambda}{n}e^{-t/\lambda} e^{-\theta(n-1)/\lambda}\quad,\,\forall\,\theta$$

Differentiating both sides of this equation wrt $\theta$, we get

$$g(\theta)e^{-n\theta/\lambda}=\frac{n-1}{n}e^{-t/\lambda}e^{-(n-1)\theta/\lambda}$$

You can now finally write down the UMVUE $g(X_{(1)})$.