Finding an unbiased estimator of$ e^{−2λ}$ for Poisson distribution

Question

The random variable X has a Poisson distribution with unknown mean λ, where 0 < λ < ∞.

Based on a single X , Is it possible to calculate an unbiased estimator of $ e^{−2λ}$ ?

I have taken an Indicator random variable as follows :-

$I_A = P(X=0) \qquad\text{if } X=0\\ \,\,\,\,\,\,= 0 \quad\qquad\qquad\text{if } X\neq0 $

Then $E(I_A) = e^{-2\lambda}$

However I am not sure if this is correct .IF This is not correct kindly help me out.

Thanks.

Ps: I have found a similar question , :Finding an unbiased estimator of $e^{-2\lambda}$ for Poisson distribution

There is only one unbiased estimator of $e^{-2 \lambda}$ from a single observation, namely $(-1)^X$, and it is not a particularly sensible estimator as it only gives the values $+1$ or $-1$ when in fact $0 \lt e^{-2 \lambda} \lt 1$ unless $\lambda=0$. See https://en.wikipedia.org/wiki/Bias_of_an_estimator#Estimating_a_Poisson_probability — Henry, Sep 10 '18 at 07:43

score 2 · Accepted Answer · edited Sep 11 '18 at 00:08

Your proposed estimator is not $I_A$ for any $A$. If it was, it should return $1$ when $X\in A$ and $0$ otherwise.

Moreover, it is also wrong to say that $$\hat \theta =\left\{\begin{matrix}P(X=0)&X=0,\\0&X\neq0,\\\end{matrix}\right.$$ is an estimator for $\theta=e^{-2\lambda}$, since $P(X=0)=e^{-\lambda}=\sqrt\theta$. Your proposed estimator for $\theta$ can be anything BUT a function that depends on $\theta$ itself.

Finding an unbiased estimator of$ e^{−2λ}$ for Poisson distribution

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