Application Problem of Expected Value of Posterior Distribution I am trying to understand the following:
Suppose that the number of people who visit the grocery store on any given day is Poisson($\lambda$) and the parameter of the Poisson distributed has a $\Gamma(3,15)$ distribution.  On Monday 62 people visited the grocery store. Derive the posterior distribution $P(\Lambda\mid\vec{X}=\vec{x})$, where $\vec{X}=(X_1,X_2,\ldots,X_n)$ and $\vec{x}$ is an observed sample from the distribution of $X$.  Then compute the posterior mean using the information provided.  Note the sum of iid Poisson r.v. is also Poisson.  
I computed the posterior distribution for a general poisson parameter with $\Gamma(a,b)$.  I found this to be $\Gamma\left(a+n,1+\frac{1}{b}\right).$  Note, the shape and scale parameterization of $\Gamma$ was used. I don't understand Why I need the fact that the sum of iid poisson r.v.'s is also poisson. Can't I just substitute $a=3$, $b=15$, $n=62$ into my posterior mean formula?
 A: For a Poisson likelihood, the gamma distribution is a conjugate prior, so the posterior is also gamma, as you have found.  This lets you calculate the posterior in two different ways:  one is to compute $f_{\Lambda}(\lambda \mid x)$ for a single observation, recognize this as gamma with updated hyperparameters, and then use an inductive argument to get $f_{\Lambda}(\lambda \mid \boldsymbol x)$ for $\boldsymbol x = (x_1, x_2, \ldots, x_n)$ for any number of observations.  Alternatively, you could note that $n \bar x = \sum x = x_1 + x_2 + \cdots + x_n$ is itself Poisson with parameter $n \lambda$, hence the posterior for $\lambda$ satisfies $$f_{\Lambda}(\lambda \mid n\bar x) = f_{\Lambda}(\lambda \mid \boldsymbol x).$$  Explicitly, $$f_{\Lambda}(\lambda \mid x) \propto \Pr[X = x \mid \lambda] f_{\Lambda}(\lambda) = e^{-\lambda} \frac{\lambda^x}{x!} \frac{b^a \lambda^{a-1} e^{-b \lambda}}{\Gamma(a)} \propto \lambda^{a+x-1} e^{-(b+1)\lambda},$$ which is clearly proportional to a gamma density with shape $a^* = a+x$ and rate $b^* = b+1$; thus for $n$ iid observations, we can extend this to get $$f_{\Lambda}(\lambda \mid \boldsymbol x) \propto \lambda^{a + \sum x - 1} e^{-(b+n) \lambda},$$ or $$\Lambda \mid \boldsymbol x \sim \operatorname{Gamma}(a + n\bar x, b + n).$$  But we could have also gotten this by writing $$f_{\Lambda}(\lambda \mid n\bar x) \propto \Pr[{\textstyle\sum} X = n\bar x \mid \lambda]f_\Lambda(\lambda) = e^{-n \lambda} \frac{(n \lambda)^{n \bar x}}{(n \bar x)!} \frac{b^a \lambda^{a-1} e^{-b \lambda}}{\Gamma(a)} \propto \lambda^{a + n\bar x - 1} e^{-(b+n)\lambda}.$$  The result is the same.
