# UMVUE of Poisson distribution truncated at zero

Suppose that $X_1,\ldots,X_n$ are $n$ i.i.d. random variables from the Poisson distribution truncated on the left at $0.$ Find the UMVUE of $P(X_1 =1).$

I am trying to do it by computing the expectation of a function $h(T)$ of my statistic, which is just the sum of $X_i$, and this expectation is equal to $P(X_1 =1),$ so I want to find $h(T),$ which will be the UMVUE, since $T$ is minimal complete sufficient statistic.

The problem is that I obtain a big double sum in this expectation and I don't know how to deal with it.

I have also tried to compute the MLE, but I don't know how to obtain it in this case.

• By "Poisson distribution truncated on the left at $0,$" I surmise that you mean the conditional distribution given that it's not $0.$ Is that right? – Michael Hardy Mar 8 '18 at 22:01
• Yes, that is what I mean. – Irene Gil Mar 8 '18 at 23:05
• You wrote "I am trying to do it by computing the expectation of a function $h(T)$ of my statistic, which is just the sum of $X_i,$ and this expectation is equal to $P(X_1=1), [\ldots]$" By "the expectation of a function $h(T)$ of my statistic", did you mean "of my statistic $T$", i.e. the statistic in question is $T$? Or by "my statistic" did you mean $h(T)$? If the thing you're trying to estimate is $P(X_1=1),$ then I think you should start with $$T=\begin{cases} 1 & \text{if } X_1 = 1, \\ 0 & \text{otherwise,} \end{cases}$$ and then seek $\operatorname E(T\mid X_1+\cdots+X_n). \qquad$ – Michael Hardy Mar 8 '18 at 23:05
• Yes, that is one option, the problem I find here is that I need the distribution of the sum of Poisson truncated random variables, which is horrible and I don´t know how to simpify the expression I get. – Irene Gil Mar 8 '18 at 23:21
• ok, I'll probably come back to this in a couple of hours. – Michael Hardy Mar 9 '18 at 0:56

I am surmising that by "Poisson distribution truncated on the left at $0,$" you mean the conditional distribution given that it's not $0.$ Thus one has $$\Pr(X_1 = x) = \frac{\lambda^x e^{-\lambda}/(x!)}{\Pr(X_1\ge 1)} = \frac{\lambda^x e^{-\lambda}/(x!)}{1 - e^{-\lambda}} = \frac{\lambda^x}{x!(e^\lambda - 1)}.$$ Thus the likelihood is $$L(\lambda) = \frac{\lambda^{x_1+\cdots+x_n}}{(e^\lambda -1)^n} \times \text{constant}$$ (where "constant" means not depending on $\lambda$), and then we have $$\ell(\lambda) = \log L(\lambda) = (x_1+\cdots+x_n)\log\lambda - n\log(e^\lambda - 1) + \text{constant}$$ $$\ell\,'(\lambda) = \frac {x_1+\cdots+x_n} \lambda - \frac{ne^\lambda}{e^\lambda - 1}.$$ I'm not sure whether finding a zero of that function can be done by using Lambert's W, but you can use numerical methods when you know $(x_1+\cdots+x_n)/n.$

I'm going to ask for some clarification of the other parts of your question before posting more here.

• I have to do it analitically, not with numerical methods. – Irene Gil Mar 8 '18 at 23:02
• I have tried with the estimator: Indicator of ${X_1}$, and then expectation of this estimator conditional by my complete minimal sufficient statistic. – Irene Gil Mar 8 '18 at 23:03
• Or other approach could be to find an estimator which is a function of the statistic, but I don`t know which estimator is that one. – Irene Gil Mar 8 '18 at 23:04
• Let $\displaystyle T = \begin{cases} 1 & \text{if } X_1=1, \\ 0 & \text{otherwise.} \end{cases}$ Then $\operatorname E(T) = \Pr(X_1=1)$ is the parameter to be estimated. If $X_1+\cdots+X_n$ is complete and sufficient, then the UMVUE is $\operatorname E(T\mid X_1+\cdots+X_n).$ We have$:\,\qquad$ – Michael Hardy Mar 9 '18 at 3:24
• We have \begin{align} & \operatorname E(T\mid X_1+\cdots+X_n=x) \\ \\ = {} & \operatorname \Pr(T=1 \mid X_1+\cdots+X_n=x) \\ \\ = {} & \Pr(X_1=1 \mid X_1 + \cdots +X_n=x) \\ \\ = {} & \frac{\Pr(X_1=1\ \&\ X_1+\cdots+X_n=x)}{\Pr(X_1+\cdots+X_n=x)} \\ \\ = {} & \frac{\Pr(X_1=1\ \&\ X_2+\cdots+X_n=x-1)}{\Pr(X_1+\cdots+X_n=x)} \\ \\ = {} & \frac{\Pr(X_1=1) \cdot \Pr(X_2+\cdots+X_n=x-1)}{\Pr(X_1+\cdots+X_n=x)} \\ \\ = {} & \frac{\lambda^x/(x!)}{e^\lambda-1} \cdot \frac{\Pr(X_2+\cdots+X_n=x-1)}{\Pr(X_1+\cdots+X_n=x)} \end{align} So we need the probability distribution of the sum. – Michael Hardy Mar 9 '18 at 3:39

To add to the final line in Michael Hardy's answer, we have that

$$E(T \mid X_1 + \cdots + X_n) = \Pr(X_1 = 1) \cdot \frac{\Pr(X_2 + \cdots + X_n = x-1)}{\Pr(X_1 + \cdots + X_n = x)}$$

Now given $X_1 + X_2 \cdots + X_n = x$ in the denominator, we can say that $X_2 + \cdots + X_n = x-X_1$.

So in the numerator, we obtain $X_2 + \cdots + X_n = x - X_1 = x -1$ , giving $X_1 = 1$.

Hence $$E(T \mid X_1 + \cdots + X_n) = \Pr(X_1 = 1) \cdot \Pr(X_1 = 1)$$

and this is easily obtainable by substituting in $x = 1$ into $\Pr(X = x) = \frac{\lambda^x}{x!(e^\lambda - 1)}$

• I hope this makes sense, someone please correct me if I'm wrong – ICE MEISTER Mar 9 '18 at 11:31
• Thank you very much to both. It really helps me! – Irene Gil Mar 10 '18 at 19:36