We will be using Neymann factorization Theorem for proving this. Let us first write down the joint density of independent Poisson random variables with the same parameter.
"f_{\\lambda}(X_1,\\ \\ldots, X_n) = e^{-n\\lambda}\\times\\frac{\\lambda^{\\sum X_i}}{\\prod_i X_i!}"
This can be written as "f_{\\lambda}(X_1,\\ \\ldots, X_n) = h(X_1,\\ \\ldots, X_n) \\cdot g(\\lambda, T(X))"
Here, T(X) = "\\overline{X}" by choosing "h(X_1,\\ \\ldots, X_n) = \\frac{1}{\\prod_i X_i!}" and "g(\\lambda, T(X)) = e^{-n\\lambda}\\times\\lambda^{n\\overline{X}}"
Now, Neymann factorization states that if we can decompose the density in the above way then T(X) which is the sample mean is sufficient for estimating the parameter "\\lambda" . Hence proved.
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