Question

Let $X_{1},X_{2},...,X_{n}$ be $n$ independent random variables, where $X_{i}$~$Poi(i\lambda )$, $i=1,2,...,n.$ Is $\sum_{i=1}^{n} X_{i}$ sufficient for $\lambda$?

Answer 1

Here, $X_{i}\sim Poi(i\lambda)\Rightarrow f(x_{i})=\frac{(i\lambda)^{x_{i}}e^{-i\lambda}}{x_{i}!}$

So, $T=\sum X_{i}\sim Poi(\lambda+2\lambda+...+n\lambda)\sim Poi(\frac{n(n+1)}{2}\lambda)$

So,$f(t)=\frac{(\frac{n(n+1)}{2}\lambda)^{t}e^{-\frac{n(n+1)}{2}\lambda}}{t!}$

Now, Joint PDF of X1,X2,.....,Xn is given by:

$f(x_{1},x_{2},....,x_{n})=\prod_{i=1}^{n}(\frac{(i\lambda)^{x_{i}}e^{-i\lambda}}{x_{i}!})=\frac{\prod_{i=1}^{n}i^{x_{i}}\lambda^{\sum x_{i}}e^{\frac{n(n+1)}{2}\lambda}}{\prod_{i=1}^{n}x_{i}!}=\frac{\prod_{i=1}^{n}i^{x_{i}}\lambda^{t}e^{\frac{n(n+1)}{2}\lambda}}{\prod_{i=1}^{n}x_{i}!}$

This can be written as:

$f(x_{1},x_{2},....,x_{n})=\frac{\prod_{i=1}^{n}i^{x_{i}}\lambda^{t}e^{\frac{n(n+1)}{2}\lambda}}{\prod_{i=1}^{n}x_{i}!}\times \frac{t!(\frac{n(n+1)}{2}\lambda)^{t}}{t!(\frac{n(n+1)}{2}\lambda)^{t}}$

$f(t).H(x_{1},x_{2},....,x_{n})$ where $H(x_{1},x_{2},....,x_{n})$ is independent of $\lambda$

Thus, by Fisher Neyman criteria, $T=\sum X_{i}$ is sufficient for $\lambda$.