Covariance between a binomial random variable and its size (number of trials) (found in the context of binomial thinning)

Question

Assume we have a random variable $X$, and we construct another random variable $Y$ to be from a binomial distribution of size $X$ and success probability $\alpha$, i.e., $Y \sim Binom(X, \alpha)$. How can you derive the covariance of $X$ and $Y$?

In the literature I am looking into (see below), it has been stated that $Cov(X, \alpha \circ X) = \alpha \cdot Var(X)$. How can I prove that?

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Background

I am studying discrete-variate time series models such as integer-valued autoregressive models (specifically, INAR(1)) (Al-Osh & Alzaid, 1987). At the core of the model lies the binomial thinning operator " $\circ$ " (first introduced by Steutel & van Harn, 1979) as defined below (rephrased from Weiß, 2018):

If $X$ is a r.v. with range $\mathbb{N}_0$ and if $\alpha \in (0;1)$, for counting series $Z_i$ (with $P(Z_i=1) = \alpha$), the r.v. $\alpha \circ X := \sum_{i=1}^{X} Z_i$ is said to arise from $X$ by binomial thinning.

It is easy to show that $(\alpha \circ X)|X \sim Binom(X, \alpha)$.

In the literature (e.g., Al-Osh & Alzaid, 1987; McKenzie, 1985; McKenzie, 2003), the autocorrelation function of INAR(1) is asserted to be $\rho(k) = \alpha^k$ (for $k \geq 0$).

As far as I could get, this is first derived by Al-Osh and Alzaid (1987) and cited afterward. In doing so (p. 265, Eq. 3.3), they substitute $Cov(X, \alpha^k \circ X)$ with $\alpha^k \cdot Var(X)$, but I am failing to derive it myself.

Answer 1

This is just addressing the narrow initial question, not the general context you cited below. The key is the Law of Total Expectation, often written as $E[X] = E[E[X|Y]]$.

A common form of covariance is: $\text{Cov}(X,Y) = E[XY] - E[X]E[Y]$.

By the Law of Total Expectation, $E[Y] = E[E[Y|X]] = E[\alpha X] = \alpha E[X]$, so the subtracted term is $\alpha E[X]^2$. Note, the "inner" expectation is over the conditional distribution of $Y$ given $X$ (i.e., the thing we understand), while "outer" expectation is just over $X$ as usual.

With both $X$ and $Y$ involved,

$E[XY] = E[E[XY | X]] = E[X E[Y|X]] = \alpha E[X^2]$. The critical step is $E[E[XY | X]]$, where the "inner" expectation is again conditioned on a specific $X$, and the outer expectation is over the variable $X$.

Thus, in total, we have

$\text{Cov}(X,Y) = E[XY] - E[X]E[Y] = \alpha E[X^2] - \alpha E[X]^2 = \alpha \text{Var}[X]$.

Hope all the steps are clear. The tricky part of working with the Law of Total Expectations is keeping the conditioning straight, sometimes it's clarifying to write it out as integrals over the given distribution.