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

  1. 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 $\pi = \frac{\mu_{X}}{mu_{X}+mu_{Y}}$.

  2. 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.

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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).

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  • $\begingroup$ I'm sorry, I have modified the content of my answer to better suit the self-study tag policy. $\endgroup$ – sjm.majewski Apr 18 '13 at 23:06
  • $\begingroup$ You could leave some of the information from your earlier answer in without carrying it quite the whole way through. In any case, I have upvoted what you have. $\endgroup$ – Glen_b Apr 18 '13 at 23:19
  • $\begingroup$ @sjm.majewski : Thanks for you reply! Using your result for $X+Y$ and the definition of conditional probability, I get: $$ P(X=k|X+Y=n) = \frac{P(X=k,X+Y=n)}{P(X+Y=n)} $$ Applying this to the above question gives: $$ P(X=k|X+Y=n) = \frac{\frac{\mu_{X}^{k}}{k!}e^{-\mu_{X}}{\frac{\mu_{Y}^{n-k}}{(n-k)!}e^{-\mu_{Y}}}}{\frac{(\mu_{X}+\mu_{Y})^{n}}{n!}e^{-(\mu_{X}+\mu_{Y})}} $$ Is this correct? If so, how do I get this in the form of the binomial distribution? $\endgroup$ – user9171 Apr 19 '13 at 13:45
  • $\begingroup$ This is correct. Now notice that the exponents cancel each other out. Then if you look closer at the thesis you'll notice you're already there ;) $\endgroup$ – sjm.majewski Apr 20 '13 at 8:00
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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.

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