Mfrhtyj

Probably a trivial mistake, but can't seem to spot it:

Assume $X_1, ldots, X_tau$ and $tau in {1, ldots, n}$ are random i.i.d variables, where $S_tau = X_1 + ldots + X_tau$ denotes the random sum.

It can be shown that the following holds:

$$Var(S_tau | tau )=tau Var(X_1)$$

However, from what I know,

$$Var(S_tau | tau) = mathbb{E}( (S_tau - mathbb{E}(S_tau|tau))^2|tau) = mathbb{E}(S_tau^2|tau) - tau^2mathbb{E}(X_1)^2$$

Here $$ mathbb{E}(S^2_tau|tau)=mathbb{E}((X_1+ldots+X_tau)^2|tau)=sum_{i,j}^{n}mathbb{E}(X_iX_j|tau)mathbb{1}_{{i,jleq tau}}=mathbb{E}(X_1^2)tau^2$$
due to independency and identical distributions.

So with my calculations I'm getting $$Var(S_tau|tau)=tau^2 Var(X_1)$$

and I'm not sure where should the squared $tau$ disappear.

I seem to be missing something, but can't spot it.

Would be grateful for any observations!

I think one oopsie is: since $X_i$ and $X_j$ are independent, $mathbb{E}[X_iX_j] = mathbb{E}[X_i]mathbb{E}[X_j] = mathbb{E}[X_1]^2$ and not $mathbb{E}[X_1^2]$! There must be another mistake, because plugging that in returns zero, but I don't see it quite yet.

There's also a much simpler proof. $var(S_tau|tau) = var(sum_{i=1}^tau X_i|tau)$. The variance of the sum of i.i.d. random variables is the sum of the variance of any random variable in the sum, so $var(sum_{i=1}^tau X_i|tau) = sum_{i=1}^tau var(X_1)$

First of all, $mathbb{E}[X_iX_j] = mathbb{E}[X_i]mathbb{E}[X_j] = (mathbb{E}[X_1])^2 ne mathbb{E}[X_1^2]$

Also, there will be $tau$ terms of $X_i^2$ and $tau(tau-1)$ terms of $X_iX_j$ in the expansion of the square. Therefore your second equation will become

$$mathbb{E}(S^2_tau|tau)=mathbb{E}((X_1+ldots+X_tau)^2|tau)=sum_{i,j}^{n}mathbb{E}(X_iX_j|tau)mathbb{1}_{{i,jleq tau}}=mathbb{E}(X_1^2)tau + tau(tau-1)(mathbb{E}[X_1])^2$$

$$Var(S_tau|tau)=mathbb{E}(X_1^2)tau + tau(tau-1)(mathbb{E}[X_1])^2 - tau^2mathbb{E}(X_1)^2$$

$$= taumathbb{E}(X_1^2) -tau(mathbb{E}[X_1])^2 $$

$$= tau(mathbb{E}(X_1^2)-(mathbb{E}[X_1])^2)$$

$$=tau Var(X_1)$$

The most succinct calculation writes the covariance of two iids using the Kronecker delta:$$operatorname{Var}S_tau=sum_{1le i,,jletau}operatorname{Cov}(X_i,,X_j)=sum_{ij}sigma^2delta_{ij}=tausigma^2.$$Your mistake, essentially, was to replace $delta_{ij}$ with $1$ throughout.

Conditional on $tau=k$, where $1leq k leq n$, $S_tau=S_k=X_1+dotsc +X_k$. For any $(X_i)$ that are IID, we have that
$$Var(X_1+dotsc+X_k)=Var(X_1)+dotsc +Var(X_k)=kVar(X_1).$$
Thus,
$$Var(X_1+dotsc +X_tau | tau =k)=Var(X_1 | tau =k)+dotsc Var(X_n |tau=k)$$
$$=Var(X_1)+dotsc +Var(X_k)=k Var(X_1),$$
and finally we obtain $Var(S_tau | tau)= tau Var(X_1)$.