EDIT: I've added some more reasoning below in the hope that it might prompt an answer from someone.
So I've been using Wald's equation to work with something that looks roughly as follows: $$\sum_{k=1}^{X_t}\mathbb{I}_k$$ where $X_t$ is a random variable which depends on the time period $t$, and $\mathbb{I}_k$ are i.i.d. indicator variables for an event that is independent of $X_t$ (for all $t$). I know that I can write $$\mathbb{E}\left(\sum_{k=1}^{X_t}\mathbb{I}_k\right) = \mathbb{E}(X_t)\mathbb{E}(\mathbb{I}_k).$$ My question is, can I do the same thing for a product of these sums, e.g. suppose that $X_t$ and $X_{t-m}$ for $m<t$ are dependent random variables. Then if $\mathbb{I}_k$ is independent of $\mathbb{I}_j$ for all $j=1,\dots X_{t-m}$, and $k=1,\dots, X_t$, is it true that $$\mathbb{E}\left[\left(\sum_{j=1}^{X_{t-m}}\mathbb{I}_j\right)\left(\sum_{k=1}^{X_t}\mathbb{I}_k\right)\right] = \mathbb{E}(X_{t-m}X_t)\mathbb{E}(\mathbb{I_j}\mathbb{I_k}).$$ I guess what I'm looking for is a version of Wald's equation for products of sums whose upper limits are dependent random variables. The reason the above equation "makes sense" to me is because there are $X_{t-m}X_t$ products of indicator variables in the expansion of the sum, so in some sense you could write it as a single sum with an upper limit of $X_{t-m}X_t$, but I'm not comfortable enough with the algebra of random variables to be certain that this is allowed.
EDIT: To add a bit more reasoning here, let's first write $$\left(\sum_{j=1}^{X_{t-m}}\mathbb{I}_j\right)\left(\sum_{k=1}^{X_t}\mathbb{I}_k\right) = \sum_{j=1}^{X_{t-m}}\sum_{k=1}^{X_t}\mathbb{I}_j\mathbb{I}_k.$$ Now, if the random variables $X_t$ and $X_{t-m}$ were to be realised as $x_t$ and $x_{t-m}$, we would have pairs $x_tx_{t-m}$ total pairs $(j,k)$, and hence $x_tx_{t-m}$ different terms of the form $\mathbb{I}_j\mathbb{I}_k$ in the double sum, so it seems natural to say that we have $X_tX_{t-m}$ different terms of this form when the $X$'s have not yet been realised. Now let's enumerate all of these pairs $(j,k)$ as $\{(j_1,k_1), (j_2,k_2),\dots, (j_{X_tX_{t-m}},k_{X_tX_{t-m}}\}$ in no particular order. Then we could write that $$\sum_{j=1}^{X_{t-m}}\sum_{k=1}^{X_t}\mathbb{I}_j\mathbb{I}_k= \sum_{l=1}^{X_tX_{t-m}}\mathbb{I}_{j_l}\mathbb{I}_{k_l},$$ and finally apply Wald's equation directly to this to get that $$\mathbb{E}\left[\left(\sum_{j=1}^{X_{t-m}}\mathbb{I}_j\right)\left(\sum_{k=1}^{X_t}\mathbb{I}_k\right)\right] = \mathbb{E}\left[\sum_{l=1}^{X_tX_{t-m}}\mathbb{I}_{j_l}\mathbb{I}_{k_l}\right] = \mathbb{E}(X_{t-m}X_t)\mathbb{E}(\mathbb{I_j}\mathbb{I_k}),$$ the desired result. The part in bold above is the part that I'm not sure is "allowed". Can you move from realised random variables to unrealised random variables in this way?
One consequence of this (for example) would be that I could use the same reasoning to compute $$\mathbb{E}\left[\left(\sum_{j=1}^{X_{t}}\mathbb{I}_j\right)\left(\sum_{k=1}^{X_t}\mathbb{I}_k\right)\right] = \mathbb{E}\left[\sum_{j=k}\mathbb{I}_j+\sum_{j\neq k}\mathbb{I}_j\mathbb{I}_k\right] = \mathbb{E}(X_t)\mathbb{E}(\mathbb{I}_j) + \mathbb{E}(X_t^2 - X_t)\mathbb{E}(\mathbb{I}_j\mathbb{I}_k).$$
Any help would be much appreciated!
Cheers
The problem with your first approach is that $\mathbb{I}_j \mathbb{I}_k$ is not identically distributed over all pairs $(j,k)$, so you cannot use Wald's equation there. Your second approach is correct for that case where $X_t$ is the same as $X_{t-m}$.
In the general case,
\begin{align} E\left[\left(\sum_{j=1}^{X_{t-m}} I_j \right)\left(\sum_{k=1}^{X_t} I_k \right)\right] &= E\left[E\left[\left(\sum_{j=1}^{X_{t-m}} I_j \right)\left(\sum_{k=1}^{X_t} I_k \right) \middle|\; X_{t-m}, X_t\right]\right] \\ &= E\left[ E\left[\sum_{j=1}^{\min\{X_{t-m}, X_t\}} I_j\;\middle|\; X_t, X_{t-m}\right] \right] + E\left[E\left[ \sum_{j=1}^{X_t} \sum_{1 \le k \le X_{t-m}; k \ne j} I_j I_k \;\middle|\; X_t, X_{t-m}\right]\right] \\ &= E[\min\{X_{t-m}, X_t\}] E[I_1] + (X_t X_{t-m} - \min\{X_t, X_{t-m}\}) E[I_1 I_2]. \end{align}