Infinite summation of modified power series (casella, berger 2nd ed [formula 7.2.22])

**shotgun1** · December 21st 2011, 08:21 AM

Hi all,

Would really appreciate if someone could assist me on the following please: Casella and berger provide an example on pg 329 where they state:

$-t_1+\log t_1\sum_{x_1=0}^{\infty}x_1\frac{e^{-t_1^{(r)}}(t_1^{(r)})^{x_1}}{x_1!}= -t_1 +t_1^{(r)}\log t_1$

I was having problems seeing how $\sum_{x_1=0}^{\infty}x_1\frac{e^{-t_1^{(r)}}(t_1^{(r)})^{x_1}}{x_1!}$ can therefore equal $t_1^{(r)}$ . Please could someone help, thanks!!

**chisigma** · December 21st 2011, 08:55 AM

Originally Posted by shotgun1

Hi all,

Would really appreciate if someone could assist me on the following please: Casella and berger provide an example on pg 329 where they state:

$-t_1+\log t_1\sum_{x_1=0}^{\infty}x_1\frac{e^{-t_1^{(r)}}(t_1^{(r)})^{x_1}}{x_1!}= -t_1 +t_1^{(r)}\log t_1$

I was having problems seeing how $\sum_{x_1=0}^{\infty}x_1\frac{e^{-t_1^{(r)}}(t_1^{(r)})^{x_1}}{x_1!}$ can therefore equal $t_1^{(r)}$ . Please could someone help, thanks!!

Writing the 'infinite sum' in a more 'standard form' You can verify that...

$\sum_{n=0}^{\infty} n\ \frac{e^{-t}\ t^{n}}{n!} = e^{-t}\ \sum_{n=1}^{\infty} \frac{t^{n}}{(n-1)!} = t\ e^{-t}\ \sum_{n=0}^{\infty} \frac{t^{n}}{n!}=t$

Marry Christmas from Serbia

$\chi$ $\sigma$

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