Question on $\operatorname{Var}(nY)$.

Question

I know there’s a simple explanation for this but it’s got me stumped. If I take the variance of $nY$, where Y is a random variable, I have $$\operatorname{Var}(nY) = \operatorname{Var}(\underbrace{Y + Y +\cdots+ Y}_{\text{$n$ times}}) = \underbrace{\operatorname{Var}(Y) + \operatorname{Var}(Y) + \cdots + \operatorname{Var}(Y)}_{\text{$n$ times}} = n \operatorname{Var}(Y).$$ But variance properties say $\operatorname{Var}(nY) = n^2\operatorname{Var}(Y)$.

Context is I’m using a Bernoulli variable Y with $E(Y) = p$ and $\operatorname{Var}(Y) = p(1-p)$

Then with $X$ = $n$ trials of $Y$, I’m practicing deriving $E(X) = np$ and $\operatorname{Var}(X) = \operatorname{Var}(nY) = n \operatorname{Var}(Y) = np(1-p).$

“$n$” cannot be squared in this case or it will be incorrect according to all sources. How do I reconcile the derivation of the variance in this case with the general property of variance of a random variable multiplied by a scalar?

Really, if you don’t need the context, my entire question is contained in the first paragraph. Why does it seem I can get two different answers for $\operatorname{Var}(nY)$.

Answer 1

It is not true that $var (X+Y)=Var (X) +var (Y)$. So you cannot write $var (nY)=Var(Y)+Var(Y)+...+Var(Y)$. The correct way to find the variance of $nY$ is to use the definition: $var (nY)=E(nY)^{2}-(E(nY))^{2}=n^{2} (EY^{2})-((nEY)^{2})=n^{2} var (Y)$.

Answer 2

Variance is not a linear operator. Rather, Covariance is a Bilinear operator: $$\mathsf{Cov}(S+T,U+V)=\mathsf{Cov}(S,U)+\mathsf{Cov}(S,V)+\mathsf{Cov}(T,U)+\mathsf{Cov}(T,V)$$

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...and so...

$$\begin{align}\mathsf {Var}(nY) &=\mathsf{Cov}(nY,nY)\\[1ex]&=n\,\mathsf{Cov}(Y,nY)\\[1ex]&=n^2\,\mathsf{Cov}(Y,Y)\\[1ex]&=n^2\,\mathsf{Var}(Y)\end{align}$$

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When $X=\sum_{k=1}^n Y_k$ where $(Y_k)$ is a sequence of independent and identically distributed Bernoulli variables. (Note, that is not identically valued.) The independence means that the covariance between two distinct members of the sequence is zero, while identical distribution means that the variance of any member of the sequence equals the variance of the first member of the sequence.$$\begin{align}\mathsf{Var}(X) &= \mathsf{Cov}(\sum_i Y_i,\sum_j Y_j)\\[1ex]&=\sum_i\sum_j\mathsf{Cov}(Y_i,Y_j)\\[1ex]&=\sum_i\mathsf{Cov}(Y_i,Y_i)+\sum_{i\neq j}\mathsf{Cov}(Y_i,Y_j)\\[1ex]& = n\,\mathsf{Var}(Y_1)+0\end{align}$$