You observe 9 cities and count the number of hurricanes that occur in each one and get the following:
Number of hurricanes: 0, 1, 2, 3, 4, 5, 6, 7, 8 Number of cities: 0, 1, 2, 2, 2, 1, 1, 0, 0
What's the poisson probability density for these results, for the following lambda values:
2, 2.5, 3, 4, 5
I have that the formula for poisson is $\frac{e^-\lambda(λ^k)}{k!}$ and λ is the mean, which for my case is $\frac{30}{9}$.
Edit: So for a mean of say, 2.5, would I do the summation of $\frac{e^-2.5(2.5^{k_i})}{k_i!}$ where k is, as described in the answer below, 1,2,2,3,3,4,4,5, and 6? How would I go about getting the Poisson Probability density for 2.5, specifically?
We have the following random sample of $X$ which is assumed to have $\mathrm{Pois}(\lambda)$ distribution for some $\lambda>0$:
$$x = (1,2,2,3,3,4,4,5,6) $$
Since $\mathbb E[X]=\lambda$, using the method of moments yields
$$\lambda = \bar x = \frac1n\sum_{i=1}^n x_i = \frac{10}3. $$
The maximum likelihood estimate is $$\hat\lambda := \arg\max\left\{\prod_{i=1}^n f(x_i;\lambda):\lambda\in(0,\infty) \right\} $$ where $$ f(x_i;\lambda) = e^{-\lambda}\frac{\lambda^{x_i}}{x_i!}.$$ Since the logarithm is an increasing function, it is equivalent to maximize the log-likelihood function: $$\ell(\lambda) = \log\left(\prod_{i=1}^n e^{-\lambda}\frac{\lambda^{x_i}}{x_i!}\right) = -n\lambda +\log\lambda\sum_{i=1}^n x_i +\sum_{i=1}^n \log x_i. $$ Differentiating with respect to $\lambda$ yields $$\frac{\mathsf d \ell(\lambda)}{\mathsf d\lambda} = -n+\frac1\lambda\sum_{i=1}^n x_i.$$ It is clear that the derivative is equal to zero if and only if $\lambda = \bar x$. Moreover, the second derivative is $$\frac{\mathsf d^2 \ell(\lambda)}{\mathsf d\lambda^2} =-\frac1{\lambda^2}\sum_{i=1}^n x_i < 0$$ for all $\lambda$, so $\hat\lambda = \bar x$ is indeed the MLE.
I don't know why the question asks to fit Poisson distributions with means $2$, $\frac52$, and $3$, but clearly those are biased estimators...