EM algorithm example 7.2.19 of Casella & Berger Statistical Inference I have been looking the Expectation Maximization example 7.2.19 in Casella & Berger (Statistical Inference) on Page 328. Here the $Y_i$ ~Poisson $(\beta\tau_i)$ and $X_i$ ~ Poisson ($\tau_i$)

I am just wondering if there is a mistake in the equation just at the top or beginning of P329. We have a term on P329 which is $\sum[\tau_i x_ilog \tau_i]$ which to me should have been $\sum [-\tau_i + x_ilog \tau_i]$. I see the same problem for the third term in the first set of parentheses. 

Is there something I am missing here if it is correct. Please help!
 A: It's a typo, in $2^{nd}$ and $3^{rd}$ terms of the first parenthesis on page 329.
If you continue, you will notice that the $3^{rd}$ term is worked properly as
\begin{eqnarray}
-\tau_1\underbrace{\sum_{x_1=0}^{\infty}\frac{e^{-\tau_1^{(r)}}(\tau_1^{(r)})^{x_1}}{x_1!}}_{\sum_{x_1=0}^{\infty}f_{X_1}(x_1)=1}
+log(\tau_1)
\color{blue}{\underbrace{\color{black}{\sum_{x_1=0}^\infty \frac{x_1\ e^{-\tau_1^{(r)}}(\tau_1^{(r)})^{x_1}}{x_1!}
}}_{\color{red}{\text{E}(X_1)=\tau_1^{(r)}, \text{ where } X_1 \sim Poisson(\tau_1^{(r)})}}} &=& -\tau_1+\tau_1^{(r)}log(\tau_1)
\end{eqnarray}
Which result is in (7.2.22).
A: I also believe it's a typo. 
$\sum[\tau_i x_ilog \tau_i]$ should be  $\sum [-\tau_i + x_ilog \tau_i]$ .
Just like the authors said, the last equality grouped terms involving $\beta$ and $\tau_i$ and terms that do not involve them. 
And the third term has the same problem. Otherwise, (7.2.22) doesn't work.
Maybe the authors have corrected them in the newest printing. I didn't see those typos in the errata. http://www.stat.ufl.edu/archived/casella/class/errata7.pdf
