In the context of a Poisson Distribution I have to compute the mean $\lambda$ from a set of empirical data.
$ X_{i} = [2, 0, 4, 1 ,\ ... ],\ hours = Y_{i} =[3.7, 6.1, 2, 13.2,\ ...], $
I am not certain how to find this particular lambda : Should it be
$\lambda = \frac{\sum_{i} X_{i}*Y_{i}}{\sum_{i}Y_{i}} $
Or simply :
$\lambda = \frac{\sum_{i} X_{i}}{\sum_{i}Y_{i}} $ ?
And what would be the intuition behind it ?
$X_{i}$ describes the number of sightings of a particular animal during an expedition, and $Y_{i}$ the time spent in the forest during the according expedition.
Thank you :)
Right now it is unclear wheter the time Y is random or fixed values, but let's assume for simplicity that the time spent is not random, hence i will write $y_i$, instead of $Y_i$.
If the amount of animals spotted in 1 hour is $poisson(\lambda)$ distributed, then the amount spotted in $y_i$ hours would be $X_i \sim poisson(y_i\lambda)$. In particular $E[\frac{X_i}{y_i}]=\lambda$, so a $possible$ unbiased estimator could be $$\frac{1}{n}\sum_{i}^n \frac{X_i}{y_i}$$ it may however be a bad estimator. Let us compute the $maximum$ $likelihood$ estimator: The log-likelihood function is \begin{align*} l(x,\lambda) &= \log(\Pi_{i=1}^n P(X_i = x_i \: | \: \lambda)) \\ &= \sum_{i=1}^n \log( \frac{(y_i\lambda)^{x_i}}{x_i!}e^{-\lambda y_i}) \\ &= \sum_{i=1}^n x_i(\log(y_i)+\log(\lambda)) - \lambda\sum_{i=1}^ny_i -\sum_{i=1}^n\log(x_i!) \end{align*} Differentiating with respect to $\lambda$ gives $$\frac{d}{d\lambda}l(x,\lambda)= \frac{\sum_{i=1}^n x_i}{\lambda} - \sum_{i=1}^n y_i $$ Setting the derivative to 0 gives us the maximum likelihood estimator: $$\hat{\lambda} = \frac{\sum x_i}{\sum y_i} $$ This is the estimate you want to use. If you are not familiar with maximum likelihood, it can be seen as the estimate which maximizes the likelihood of observing the given data (hence the name).