Finding $E[X^2]$ for $X \sim Bin(25,0.61)$

Question

I got something rather new and I just wanted to make sure my way of thinking in this field is fine. Suppose $$X\sim Bin(25,0.61)$$ and we are asked to find: $E[X^2]$. So basically I treat this binomial variable as a sum of 25 Bernoulli variables where: $$X_i=\begin{Bmatrix} 1 & p=0.61\\ 0 & p=0.39 \end{Bmatrix}$$ So if I got it right, the fact I've got a binomial variable that I split into Bernoulli variables kind of gives me the privellege to "ignore" the sqaure part and: $$E[X^2] = E[X] = \sum_{i=1}^{25}{}E[X_i]=\sum_{i=1}^{25}0.61\times1 = np = 25\times 0.61 = 15.25$$ Is that a valid move / assumption since we're talking about a binomial variable?

Answer 1

Your calculation is not correct.

$X\sim\text{Binomial}(n,p)\iff X=\displaystyle\sum_{i=1}^{n}Y_i,$ where $Y_i\stackrel{iid}{\sim} \text{Bernoulli}(p).$ In this case $n=25$ and $p=0.61.$

$X\neq X^2$ since $X$ could be neither $0$ nor $1.$ The Bernoulli random variables $Y_i$ are $0$ or $1,$ and so $Y_i=Y_i^2.$

Note that $X^2=\displaystyle\sum_{i=1}^{n}Y_i^2+\underset{i\neq j}{\displaystyle\sum_{i=1}^{n}\sum_{j=1}^{n}}Y_iY_j=\displaystyle\sum_{i=1}^{n}Y_i+\underset{i\neq j}{\displaystyle\sum_{i=1}^{n}\sum_{j=1}^{n}}Y_iY_j.$

Now $Y_i$ are iid, $\mathbb{E}(Y_i)=p,$ and so $\underset{i\neq j}{\displaystyle\sum_{i=1}^{n}\sum_{j=1}^{n}}\mathbb{E}(Y_iY_j)=n(n-1)\mathbb{E}(Y_i)\mathbb{E}(Y_j)=n(n-1)p^2,$ which gives

$$\mathbb{E}(X^2)=np+n(n-1)p^2=238.51$$

Answer 2

We cannot simply ignore the square, but you have a part of the answer. If we have $X = \sum_{i=1}^{25}X_{i}$ where $X_{i} \sim Ber(0.61)$, then

$$X^2 = (\sum_{i=1}^{25}X_{i})^2 = \sum_{i=1}^{25}X_{i}^2 + 2\sum_{1 \leq i < j \leq 25} X_{i}X_{j}.$$

By linearity of expectation, we have

$$E(X^2) = \sum_{i=1}^{25}EX_{i}^2 + 2 \sum_{1 \leq i < j \leq 25} E(X_i X_j).$$

Since $X_{i}$ is Bernoulli, we can "ignore" the square for $EX_i^2$. So

$$\sum_{i=1}^{25}EX_{i}^2 = \sum_{i=1}^{25}EX_{i} = 15.25.$$

But you still need to calculate the second part above. Since the $X_i$ are independent, we have $E(X_i X_j) = EX_i EX_j$. How to finish?