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I have to find UMVUE for $exp(-k*a)$ where X ~ Exponential(a); k is a positive real number.

I tried it using Lehmann-Scheffe theorem. Since, T = $sum(xi) (i = 1,..,n)$ is complete sufficient statistic for a, we need to find g(T) such that,

E(g(T)) = $exp(-k*a)$

I further equated the expectation of g(t) to $exp(-k*a)$ using the fact T ~ $Gamma(n,a)$, but I am unable to proceed any futher, any help is appreciated.

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  • $\begingroup$ I don't think so, because my question is a continuous case. $\endgroup$
    – Kcd
    Nov 6, 2021 at 8:52
  • $\begingroup$ Or stats.stackexchange.com/q/374010/119261 $\endgroup$ Nov 6, 2021 at 10:02
  • $\begingroup$ I understood the method used in the question for UMVUE of P(X<2). We can generalize that for k as 1 - (1 - k/T)^(n-1). But, from this can we infer that the UMVUE I require would be 1 - (1 - (1 - k/T)^(n-1)) ? If yes, please explain how? $\endgroup$
    – Kcd
    Nov 6, 2021 at 11:39
  • 2
    $\begingroup$ Here is a tutorial for typesetting math. Look at the solutions again, including the questions. $\endgroup$ Nov 6, 2021 at 13:14

1 Answer 1

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Below is a failed approach:

You could try to compute the maximum likelihood estimator of $t_k= e^{k\lambda}$. We know that the estimator for $1/\lambda$ is $\frac{1}{n}\sum_{k=1}^n{X_k} = S_n/n$ so let's plug that in:

$$\hat{t}_{k,MLE} = e^{-kn/S_n}$$

The distribution of $S_n$ is an Erlang distribution

$$S_n \sim \frac{\lambda^n s^{n-1}e^{-\lambda s}}{(n-1)!}$$

when we apply the transform $y = e^{-kn/s}$ or $s = -kn/\log(y)$ we get:

$$Y \sim \frac{\lambda^n (-kn/\log(y))^{n-1}e^{\lambda kn/\log(y)}}{(n-1)!} \frac{ds}{dy} = \frac{\lambda^n (-kn)^{n}}{(n-1)!} \frac{\left(e^{\lambda kn}\right)^{1/\log(y)}}{y\log(y)^{n+1}}$$

Then you need to compute $E[Y]$ to see the bias of $e^{-kn/Sn}$ to correct for it. This will be some function of $\lambda$, $k$ and $n$ but at this point, it doesn't look pretty, and this approach might not be fruitful.

Small Improvement

We know that $S_n$ follows a gamma distribution so $X = 1/S_n$ follows an inverse gamma distribution. Which has characteristic function

$$\varphi(t) = E[e^{itX}] = \frac{2(-i\beta t)^{\alpha/2}}{\Gamma(\alpha)}K_{\alpha}(\sqrt{-4i\beta t})$$

with $\alpha = n$ and $\beta = \lambda$

and

$$E[e^{-kn/S_n}] = E[e^{-kn X}] = \varphi(ikn) = \frac{2(\lambda kn)^{n/2}}{\Gamma(n)}K_{n}(\sqrt{4\lambda kn})$$

which still doesn't look pretty. The modified Bessel function of the second kind $K_{\alpha}$ does approach some exponential term such that you get a term $e^{k\lambda}$ but there is also the term $(k\lambda)^{n/2}$, so we do not get a bias that can be easily resolved by some scaling factor.

Check with a simulation

The R-code below checks the formula above. The mean of the estimator from a simulation corresponds well with the computed mean based on the above formula.

k = 2
n = 5
lambda = 3

### comuted expectation
mu_comp = 2*(k*n*lambda)^(n/2) * besselK(sqrt(4*k*n*lambda),n) / factorial(n-1)

### simulated expectation
simulate = function() {
  x = rexp(n,lambda)
  exp(-k*n/sum(x))
}

set.seed(1)
### simulated expectation of estimator
### 0.007904434
y = replicate(10^5, simulate())
mean(y)

### computed expectation of estimator
### 0.007909603
mu_comp

### true value of exp(-k*lambda)
### 0.002478752
exp(-k*lambda)

Approximation

For large $n$ the sum $S_k$ will follow approximately a normal distribution, and we can estimate the distribution of the estimator $\hat{t}_k = e^{-kn/S_n}$ with the delta method by linearizing the function around the mean of $S_n$ (which we call $\mu_{S_n}$)

$$\begin{array}{} \hat{t}_k &=& e^{-kn/S_n} \\ &\approx& e^{-kn/\mu_{S_n}} + (S_n - \mu_{S_n}) \frac{kn e^{-kn/\mu_{S_n}}}{\mu_{S_n}^2} + \frac{1}{2}(S_n - \mu_{S_n})^2 \frac{kn e^{-kn/\mu_{S_n}} (kn-2\mu_{S_n})}{\mu_{S_n}^4} \\ E[\hat{t}_k]&\approx &e^{-kn/\mu_{S_n}} E \left[1 + \frac{1}{2} (S_n - \mu_{S_n})^2 \frac{kn (kn-2\mu_{S_n})}{\mu_{S_n}^4} \right] \end{array}$$

filling in $\mu_{S_n} = n/\lambda$ and $E[(S_n - \mu_{S_n})^2] = n/\lambda^2$

$$\begin{array}{} E[\hat{t}_k] &\approx & e^{-k\lambda} E \left[1 + \frac{1}{2} n/\lambda^2 \frac{kn (kn-2n/\lambda)}{(n/\lambda)^4} \right] \\ &\approx & e^{-k\lambda} E \left[1 + \frac{\frac{1}{2}k^2 \lambda^2-k\lambda}{n} \right] \end{array}$$

### delta estimate expectation of estimator
### 0.008427757
exp(-k*lambda) * (1+(0.5*k^2*lambda^2-k*lambda)/n)

We can estimate the $\lambda^2$ and $\lambda$ with $\frac{(n-1)(n-2)}{S_k^2}$ and $\frac{n-1}{S_k}$, then an approximate estimator can be

$$\hat{t}_{k,corrected} = \frac{e^{-kn/S_n}}{1+{\frac{0.5 k^2 (n-1)(n-2)}{n S_k^2}- \frac{k (n-1)}{n S_k}}} $$

### simulated expectation of corrected estimator
simulate2 = function() {
  x = rexp(n,lambda)
  Sn = sum(x)
  estlambda = (n-1)/Sn
  estlambda2 = (n-1)*(n-2)/Sn^2
  exp(-k*n/Sn) / (1+(0.5*k^2*estlambda2-k*estlambda)/n)
}

set.seed(1)
### simulated expectation of corrected estimator
### 0.007321175
y = replicate(10^5, simulate2())
mean(y)

This first-order bias-corrected estimator does only slightly better.

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