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In my book, it says:

Independent random variables $X_1, X_2, \dots, X_n$ are modeled by a Poisson distribution with mean $\lambda > 0$. The likelihood for $\lambda$ based on data $\mathbf{x}=(x_1,x_2,\dots\,x_n)^T$ is $$L(\lambda)=\prod_{i=1}^n \frac{\lambda^{x_i}e^{-\lambda}}{x_i!}$$ $$=k \lambda^{n \bar{x}}e^{-n\lambda}$$ where $k$ is a constant.

At first I was a bit confused by this, but I think I get it now. The $e^{-n\lambda}$ bit is easy. For the $\lambda^{n \bar{x}}$ part I think they had $\lambda^{x_1+x_2+\dots,x_n}=\lambda^{\sum{x_i}}$ so $\bar{x}=\frac{\sum{x_i}}{n}$ and $\sum{x_i}=n \bar{x}$ and then $k= \frac{1}{\prod_{i=1}^nx_i!}$ which doesn't depend on $\mathbf{\lambda}$. Have I missed anything ?

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    $\begingroup$ No, it seems ok to me. Recall that $\prod_i(a_i\times b_i \times...)=\prod_i(a_i)\times \prod_i(b_i)...$, which is exactly what you did. $\endgroup$
    – Néstor
    Jul 9 '12 at 7:39
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    $\begingroup$ I agree. You have it right. $\endgroup$
    – Michael R. Chernick
    Jul 9 '12 at 9:59
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Your interpretation of the algebra is correct. It is sometimes easier to express these things more succinctly using the proportionality relation $\propto$. Using this relation you can drop $k$ and state that:

$$p( \boldsymbol{x} | \lambda) = \prod_{i=1}^n \frac{\lambda^{x_i}e^{-\lambda}}{x_i!} \propto \lambda^{n \bar{x}}e^{-n\lambda} = L_\boldsymbol{x}(\lambda).$$

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