Compute $\mathbb{E}[X]$ and $Var[X]$ for $\lambda=1$

Question

Let X be a discrete random variable such that ($\lambda\geq0)$ $$P(X=k)=\frac{\lambda^k}{k!}e^{-\lambda}\quad\text{for}\;k\in\{0,1,...\} $$

1. Check that $\displaystyle{\sum^\infty_{k=0}P(X=k)=1}$

2. Compute $\mathbb{E}[X]$ for $\lambda=1$

3. Compute Var$[X]$ for $\lambda=1$

My attempt:

For the first part, we can use the fact that the taylor expansion of $e^x$ is $e^x=\displaystyle{\sum_{k=1}^\infty\frac{x^k}{k!}}\;\forall\;x\in\mathbb{R}$. Then, $$\displaystyle{\sum^\infty_{k=0}P(X=k)=\sum^\infty_{k=0}\frac{\lambda^k}{k!}e^{-\lambda}=e^\lambda\cdot e^{-\lambda}=e^0=1}$$

For the second part, since we know that this is a valid p.d.f (from the first part), we can use the formula $\mathbb{E}[X]=\displaystyle{\sum_{k\in X}kp(k)}$. Then, $$\mathbb{E}[X]=\sum k\cdot\frac{1^k}{k!}e^{-1}=\sum \frac{1^k}{(k-1)!} e^{-1}=1$$

This is where I am stuck. I don't know if I am finding the expectation correctly because the result seems strange.

Once I am able to find this, I can find the third part by using the formula $\text{Var}[X]=\mathbb{E}[X^2]-\mathbb{E}[X]^2$ to find the variance.

I also have a question about finding $\mathbb{E}[X^2]$. Is it $\displaystyle{\sum_{k\in X}(kp(k))^2}$ or $\displaystyle{\sum_{k\in X} k^2p(k^2)}$?

Answer 1

You have $\displaystyle \sum_{x=0}^\infty \frac{\lambda^x e^{-\lambda}}{x!} = 1.$ Then $$ \operatorname E(X) = \sum_{x=0}^\infty x \Pr(X=x) = \sum_{x=0}^\infty x\cdot \frac{\lambda^x e^{-\lambda}}{x!} $$ Now notice that in the line above

• I am carefully distinguishing between two different things, denoted capital $X$ and lower-case $x$ (and I partially cleaned up the notation in the question);

• When $x\ge1$ we have $x\cdot\dfrac 1 {x!} = \dfrac 1 {(x-1)!},$ but when $x=0$ that cannot be done. And need not be done, since that term is $0.$

Since the $x=0$ term of the last sum is $0$ we can drop it and get $\displaystyle\sum_{x=1}^\infty x\cdot\frac{\lambda^x e^{-\lambda}}{x!}$ (starting at $x=1,$ not at $x=0$). And then we can do the cancelation noted above and get $\displaystyle \sum_{x=1}^\infty \frac{\lambda^x e^{-\lambda}}{(x-1)!}.$ And this is equal to $$ \lambda\sum_{x=1}^\infty \frac{\lambda^{x-1} e^{-\lambda}}{(x-1)!}. $$ If we let $y=x-1$ then we notice that as $x$ goes from $1$ to $\infty$ then $y$ goes from $0$ to $\infty,$ and the sum becomes $$ \lambda\sum_{x=1}^\infty \frac{\lambda^y e^{-\lambda}}{y!} $$ and that sum is equal to $1,$ so the expected value is $\lambda.$

Maybe the simplest way to find $\operatorname E(X^2),$ and thus to get $\operatorname{var}(X) = \operatorname E(X^2) - \big(\operatorname E(X) \big)^2,$ is to find $\operatorname E(X(X-1)) = \operatorname E(X^2) - \operatorname E(X),$ as follows: $$ \operatorname E(X(X-1)) = \sum_{x=0}^\infty x(x-1)\cdot\frac{\lambda^x e^{-\lambda}}{x!}. $$ Now notice that the first two terms, with $x=0$ and $x=1,$ are $0.$ And so as above we get $$ \lambda^2 \sum_{x=2}^\infty \frac{\lambda^{x-2} e^{-\lambda}}{(x-2)!} = \lambda^2 \sum_{y=0}^\infty \frac{\lambda^y e^{-\lambda}}{y!} $$ where this time $y=x-2.$ Again the sum is $1$ and we have $\operatorname E(X(X-1)) = \lambda^2.$ Therefore $\operatorname E(X^2) = \lambda^2+\lambda,$ and so $\operatorname{var}(X) = \lambda.$

Answer 2

The key is to always be aware of the supports for your series.   Then you may notice that the first term vanishes, and a change of variables will do the rest.

$$\begin{align} \mathsf E(X) &= \sum_{k=0}^\infty k\cdot\dfrac{1^k\mathrm e^{-1}}{k!} \\[1ex]&=\mathrm e^{-1}\left(0\cdot\dfrac{1}{0!}+\sum_{k=1}^\infty\dfrac{1}{(k-1)!}\right)\\[1ex]&=\mathrm e^{-1}\sum_{j=0}^\infty\dfrac{1}{j!}\\[1ex]&= \mathrm e^{-1}\,\mathrm e^1\\[1ex]&=1 \end{align}$$