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Let $X_n$ be the numbers of job applications at a company in the year $1900+n,n\in\mathbb N$. Let $(X_n)_{n\in\mathbb N}$ be a sequence of independent, identically distributed random variables with the Poisson ($\lambda$) distribution, where $\lambda\in[1,144]$. Give an approximation of $\lambda$ based on $(X_n)_{n=1}^{100}$ by using a law of large numbers (= weak law or strong law).

The strong law of large numbers gives $$ \operatorname{lim}_{n\to\infty} \frac{1}{n}S_n=\operatorname{lim}_{n\to\infty}\frac{1}{n}(X_1+\dots+X_n)=\mu\quad\text{in mean square.} $$ This means that $$ \mathbb E([\frac{1}{n}S_n-\mu]^2)\to0\quad\text{as }n\to\infty. $$ We know that the mean of this distribution is $\lambda$, so we can write $$ \operatorname{lim}_{n\to\infty}\mathbb E([\frac{1}{n}S_n-\lambda]^2)=0. $$ As we are going to give an approximation, we're going to use $$ \mathbb E([\frac{1}{100}S_{100}-\lambda]^2)\approx0. $$ This means that $\mathbb P(\frac{1}{!00}S_{100}=\lambda)\approx 1$, but I don't know how to continue from here on.

I see that $\mathbb E([\frac{1}{100}S_{100}-\lambda]^2)=\operatorname{var}(\frac{1}{100}S_{100})=\frac{1}{10^4}\cdot100\operatorname{var}(X)=\frac{\lambda}{10^4}100=0.$

But this would mean $\lambda$ is zero. So what is my mistake?

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  • $\begingroup$ I am not sure about your $var(\frac{1}{100}S_{100})=\frac{1}{10^{4}}100var(X)$. Is that true? Also, I do not understand your question. What 'approximation of $\lambda$' means? $\endgroup$ – Jan Feb 3 '17 at 0:38
  • $\begingroup$ @Jan Yes it's true, because all $X_i$ are independent. You just need to give a numerical approximation of the Poisson parameter $\lambda$. $\endgroup$ – Sha Vuklia Feb 3 '17 at 0:40
  • $\begingroup$ I was not doubting $var(X_{1}+X_{2})=2var(X_{1})$, I was doubting $var(cX)=c^2var(X)$, which holds for normal but Poisson. About the parameter, I am still unclear what you are trying to get at. Are the $1900+n$ realizations, that is, your data? $\endgroup$ – Jan Feb 3 '17 at 0:44
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    $\begingroup$ @Jan $var(cX)=c^2var(X)$ holds for all random variables, which is easily proven by definition. $\endgroup$ – Blaza Feb 3 '17 at 1:15
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    $\begingroup$ @ShaVuklia You are asked to give an approximation for $\lambda$ and you know that $\frac {S_n}n \to \lambda$. Why don't you just take that as the approximation, i.e. $\hat\lambda=\frac{S_n}n$? In your work with the expectation, you actually showed that $\frac{S_n}n $ is a good estimate, as it has a low mean squared error. $\endgroup$ – Blaza Feb 3 '17 at 1:26

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