Questions on likelihood analysis Whilst studying likelihood methodologies, I've come across some results that I haven't been able to work out.


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*If $X$ and $Y$ are Poisson with means $\mu_{X}$ and $\mu_{Y}$, then the conditional distribution of $X$ given $X+Y$ is binomial with parameters $n = X+Y$ and $\pi = \frac{\mu_{X}}{mu_{X}+mu_{Y}}$.

*Assuming $y_{1}, \ldots, y_{n}$ are independent exponential outcomes with mean $\mu_{1}, \ldots, \mu_{n}$. Given $\text{log}(\mu_{i}) = \beta_{0}+\beta_{1}x_{i}$ and $\sum x_{i} = 0$, the profile likelihood of $\beta_{1}$ is $$ \text{log}(L_{p}(\beta_{1})) = -n \log \left(\sum_{i}y_{i}e^{-\beta_{1}x_{i}}\right)$$
I would include my attempts at showing these results, but I'm completely lost here. Perhaps somebody could provide some assistance.
 A: For the first problem, try to compute $P(X = k | X+Y = n)$. It is helpful to know, that the sum of two independent Poisson variables $X$, $Y$ is a Poisson variable with the mean equal to the sum of means of $X$ and $Y$ (proving this fact is a good elementary problem in itself).
A: For the second problem, begin by writing the likelihood followed by the log likelihood to get: 
\begin{align}
L(\beta_0, \beta_1) &= \prod_i\left(\mu_i  e^{-\mu_i^{-1}y_i}\right) \\
\ell(\beta_0, \beta_1) &= - \sum_i\left(\ln\mu_i + \mu_i^{-1}y_i\right) \\
&= - \sum_i\left(\beta_0+\beta_1x_i + y_ie^{-\beta_0-\beta_1x_i}\right)
\end{align}
Consider $\beta_0$ as the nuisance parameter and $\beta_1$ as the parameter of interest. To find the profile likelihood of $\beta_1$ the idea is to solve for the MLE of your nuisance parameter (which will ideally contain no terms involving $\beta_0$ - as is the case here) and plug that into the log likelihood. Doing so will give:
\begin{align}
\ell(\beta_0, \beta_1) &= - \sum_i\left(\ln\left(\sum_i y_i e^{-\beta_1x_i}\right)\right)-\beta_1\sum_ix_i
\end{align}
Now the question states that $\sum_i x_i = 0$ and so the required answer follows.
