Difficult to understand difference between the estimates on E(X) and V(X) and the estimates on variance and std.dev. on lambda-hat I'm having a very hard time to separate estimates on population values versus estimates on sample values.
I'm struggling with this exercise (not homework, self-study for my exam in introductionary statistic course): 

a) Calculate the maximum likelihood estimate on $\lambda$ and name this $\hat{\lambda}$
I did this. First I calculated the maximum likelihood estimator for $\lambda$, which is $\hat{\lambda}=\frac1n \sum_{i=0}^n k_i$ 
And the maximum likelihood estimate was $\hat{\lambda_e}=\frac{15}6=2.5$
Hopefully this is right. But now I'm asked to calculate the "estimates on $E(X)$ and $E(Y)$ and afterwards I'm asked to calculate the estimate on both the variance and standarddeviation on the estimate, $\hat{\lambda}$?
I'm really struggling to separate the two from each other, so any hint or guidance would be much appreciated. 
 A: First of all you've got the correct equation:
$$\hat{\lambda} = \frac{1}{n}\sum_{i=1}^n x_i$$ 
but I think you've misunderstood the data given in the question. The table says that there are 123 instances of $X$ taking the value 0. Similarly there are 67 instances of the value 1. Try using this to calculate $\hat{\lambda}$ (I get 0.68).
For $E(X)$ you should think about what the expectation is of a general poisson distribution, in terms of $\lambda$, then substitute your estimate of $\lambda$ to get an estimate of $E(X)$.
To find the variance of $\hat{\lambda}$, you are right that an estimator is a random variable. This particular one (as you found at the beginning) is given by:
$$\hat{\lambda} = \frac{1}{n}\sum_{i=1}^n X_i$$
where $X_i \sim Poi(\lambda)$. Since each of the $X_i$ have the same distribution, they have the same variance. You can estimate this by looking up the variance of a general Poisson($\lambda$) distribution in terms of $\lambda$ and then substituting in your estimate for $\lambda$. 
To find the variance of $\hat{\lambda}$ you will then need the following facts: 


*

*$Var(aX)=a^2Var(X)$ when $X$ is a random variable and $a$ is any number

*$Var(X + Y) = Var(X) + Var(Y)$ when $X$ and $Y$ are independent of each other. You can almost certainly assume all of your $X_i$'s are independent in this question.


The estimated standard deviation is then the square root of the estimated variance.
A: It's not clear how you got $15/6$.  The sample mean is
\begin{align}
& \frac{\overbrace{0+\cdots+0}^{\text{123 of these}}+\overbrace{1+\cdots+1}^{\text{67 of these}}+ \overbrace{2+\cdots+2}^{27}+\overbrace{3+\cdots+3}^{10}+5}{123+67+27+10+0+1} \\[10pt]
= {} & \frac{156}{228} = 0.68421\ldots
\end{align}
For the Poisson distribution the variance is the same as the mean, and you can take the statistic above to be an estimate of the mean and of the variance.
The variance of $\hat\lambda$ is
\begin{align}
& \operatorname{var}\left(\frac{X_1+\cdots+X_{228}}{228}\right)
= \frac 1 {228^2}\operatorname{var}(X_1+\cdots+X_{228}) \\[10pt]
= {} & \frac 1 {228^2} \left(\operatorname{var}(X_1)+\cdots+\operatorname{var}(X_n) \right) =\frac 1 {228^2} (\lambda+\cdots+\lambda) \\[10pt]
= {} & \frac 1 {228^2} \cdot 228\lambda = \frac \lambda {228}.
\end{align}
