A used car salesman is willing to assume the number of sales he makes, per day, is a Poisson random variable with parameter $\mu$. Over the past 30 days, he made $0$ sales on $20$ days and one or more sales on each of the remaining $10$ days. Show that the the method of moments and maximum likelihood estimations for $\mu$ is 0.41
So I'm stuck trying to figure out how you estimate the ML or MoM estimator without knowing how many cars were actually sold. For example, for the ML estimator:
$L_X(\mu) = \prod\limits_{i=1}^{30}\frac{\mu^{x_i}}{x_i!}e^{-\mu} =
\frac{\mu^{\sum{x_i}}}{\prod{x_i!}}e^{-30\mu}$
taking the log likelihood:
$K(\mu) = \sum{X_i}ln{\mu} - \sum{ln{x_i!}} - 30 \mu$
and taking the derivative w.r.t. $\mu$:
$\dfrac{\partial K}{\partial \mu} = \dfrac{\sum{X_i}}{\mu} - 30 = 0$
and thus:
$\mu = \dfrac{\sum{X_i}}{30}$
but obviously I'd need to number of sales to complete this equation. My intuition is that I'd have to somehow take into account that he didn't sell any cars for 20 days, and then sold 1 or more for the remaining 10 days, but I'm not sure how to implement that knowledge in the ML or MoM.
What was the joint probability of the result that was observed? That is to say, if $X_i \sim \operatorname{Poisson}(\mu)$ is the random number of cars sold on day $i$, and each day is independent of the others, what was the probability of observing $20$ days with $X_i = 0$ and $10$ days with $X_i > 0$? Because each $X_1, \ldots, X_{30}$ is independent and identically distributed, we can assume without loss of generality that $X_1 = \ldots = X_{20} = 0$ (the first $20$ days he sold no cars), and $X_{21}, \ldots, X_{30} > 0$ (the last $10$ days, he sold at least one car). Then the joint probability of this observation is $$\Pr[\boldsymbol X = \boldsymbol x] = \left(e^{-\mu} \frac{\mu^0}{0!}\right)^{20} \left(1 - e^{-\mu} \frac{\mu^0}{0!} \right)^{10} = e^{-20\mu} (1 - e^{-\mu})^{10},$$ where $\boldsymbol X = (X_1, X_2, \ldots, X_{30})$ is the random vector of the number of cars sold on each day, and $\boldsymbol x = (0, 0, \ldots, 0, x_{21}, \ldots, x_{30})$ is the observed vector of cars sold on each day. Note that we don't actually know $x_{21}, \ldots, x_{30}$. But we can write the joint probability because we know for each $i \in \{21, \ldots, 30\}$, the number of cars sold is at least $1$; thus the probability of this event occurring is the complement of the probability of selling zero cars that day. This is how we arrive at the above joint probability.
Consequently, the likelihood function given the sample is simply $$\mathcal L(\mu \mid \boldsymbol x) = e^{-20\mu}(1-e^{-\mu})^{10}, \quad \mu > 0.$$ The log-likelihood is $$\ell(\mu \mid \boldsymbol x) = -20 \mu + 10 \log(1-e^{-\mu}),$$ and the critical points satisfy $$0 = \frac{\partial \ell}{\partial \mu} = -20 + \frac{10e^{-\mu}}{1-e^{-\mu}},$$ or $e^{-\mu} = 2/3$, from which we can conclude that the likelihood is maximized for $$\mu = \hat\mu = \log \frac{3}{2} \approx 0.405465.$$