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I'm studying sufficient statistics and I came across this problem:

A dataset consists of independent triples $(W_i,Y_i,Z_i)$ of independent random variables with distributions as follows, $$ W_i \sim \mathcal{N}(\lambda+\tau,1), \, \, Y_i \sim \mbox{Exp}(\lambda+\mu) \, \, Z_i \sim \mbox{Poisson}(\mu+\tau), \, i=1,\ldots,n$$

(here $\mathcal{N}$ denotes normal distribution parameterized by mean and variance, and $\mbox{Expo}(\theta)$ the exponential distribution with mean $\frac{1}{\theta}$.

The question asks if there is a UMVU estimator for $\lambda$ and to prove (or disprove) it.

Although the question above is not the question I'm asking, any tips would be appreciated. (my question will come after the equation).

Now, if I did everything right, the joint density of the triplet is: \begin{align} \small{f(\mathbf{w},\mathbf{y},\mathbf{z}; \lambda, \tau,\mu) = \frac{(2 \pi)^{-\frac{n}{2}}}{\prod_i z_i!} e^{\frac{\sum_i w_i^2}{2}} \cdot \left[(\lambda+\mu)^n e^{-n(\mu+\tau)+n\frac{(\lambda +\tau)^2}{2}}\right] \cdot \, \, \quad \exp\left\{-(\lambda+\tau)\sum_i w_i -(\lambda+\mu) \sum_i y_i + \log(\mu+\tau)\cdot \sum_i z_i \right\}} \end{align}

So, this is an exponential family, and by the factorization theorem we can find that $(\sum_i W_i,\sum_i Y_i,\sum_i Z_i)$ is the vector of sufficient statistics.

Now, my questions are:

1) are they sufficient to what? The vector $(\lambda,\mu,\tau)$? The vector $(\lambda+\tau,\lambda+\mu,\mu+\tau)$?

1.1) If it's the last one, can I say that $\sum_i W_i$ is sufficient for $(\lambda +\tau)$; $\sum_i Y_i$ is sufficient for $(\lambda+\mu)$; and $\sum_i Z_i$ is sufficient for $(\mu+\tau)$? Can I make this "one-to-one" relation?

2) can I rewrite the exponential as $$\small{\exp\left\{-(\lambda+\tau)\sum_i w_i -(\lambda+\mu) \sum_i y_i + \log(\mu+\tau)\cdot \sum_i z_i \right\} = \exp\left\{-\lambda(\sum_i w_i + \sum_i y_i) -\tau \sum_i w_i -\mu \sum_i y_i + \log(\mu+\tau)\cdot \sum_i z_i \right\}}$$ and say that $(\sum_i w_i + \sum_i y_i)$ is sufficient for $\lambda$? In this case, what would be the sufficient statistics for $\tau$ and $\mu$? (this is somewhat related to 1.1)

Thanks

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    $\begingroup$ Re (1): If you know $(\lambda,\mu,\tau)$ then you know $(\lambda+\tau,\lambda+\mu,\mu+\tau)$, and vice-versa. Thus there is no content to this question. Perhaps that insight could be of some help in (2). $\endgroup$
    – whuber ♦
    Jul 8, 2016 at 13:55
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    $\begingroup$ @whuber Thanks. So, $(\sum_i W_i,\sum_i Y_i,\sum_i Z_i)$ is sufficient for any linear combination of $\lambda$, $\tau$, and $\mu$ (including $\lambda + 0 \cdot \tau + 0 \cdot \mu$), and in order for me to find an UMVUE for $\lambda$ I should look for estimators that depend on all three summations. Is that it? $\endgroup$
    – ogustavo
    Jul 9, 2016 at 22:27

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