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$\newcommand{\Cov}{\operatorname{Cov}} \newcommand{\E}{\mathbb{E}}$

I realize that $\Cov(X,Y) = \E[(X-\mu_X)(Y-\mu_Y)] = \E[\Cov(X,Y|A)] + \Cov(\E[X|A], \E[Y|A])$. But I am not sure how this is applied to functions of random variables conditioned on another random variable belonging to the same probability space.

Let $g(X,Y) = X / (X+Y)$ and $h(X,Y,Z) = (X+Y) / (X+Y+Z)$ , where $X, Y, Z$ are independent Poisson random variables with rates $\lambda_X$, $\lambda_Y$, and $\lambda_Z$, respectively, and where $X + Y + Z > 0$ and $g(X,Y) = 0$ if $Z=N$.

Suppose we know $N = X + Y + Z$.

What is the conditional covariance of $g(X,Y)$ and $h(X,Y,Z)$ given $N$ and how does one go about finding it?

Possibly this helps (?): For a known $N$, $h(X,Y,Z|N) \sim Binom(N,p) / N$ , where $p$ is the probability of success. This can be shown by recognizing $X+Y$ to be a sum of independent Poisson random variables, and thus also a Poisson random variable with rate $\lambda_X + \lambda_Y$. Conditioned on $(X+Y) + Z = N$, $(X+Y)$ is binomial with $p = (\lambda_X+\lambda_Y)/(\lambda_X + \lambda_Y + \lambda_Z)$.

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    @MichaelHardy No worries. Maybe I am missing something, but the problem solution does not seem straightforward. And I have found little online in the way of worked examples to guide the process. Is the way forward sheets of algebra, or should some sort of propagation of errors technique be used?2012-07-03

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Based on the HINT from @did, here is a solution for the covariance of $g(X,Y|N)$ and $h(X,Y,Z|N)$.

$\DeclareMathOperator \Cov {Cov}$ $\DeclareMathOperator \E {E}$ $\DeclareMathOperator \Var {Var}$ $\Cov(g(X,Y), h(X,Y,Z)|N) = \E\biggl[ g(X,Y)h(X,Y,Z)\bigg|N\biggr] - \E\biggl[g(X,Y)\bigg|N\biggr]\ \E\biggl[h(X,Y,Z)\bigg|N\biggr]$

First consider $\E[X|N]$ $\ldots$

$\E[X|X+Y+Z] = \E[X|X+W] = \frac{\lambda_X}{\lambda_X + \lambda_W} (X+W) = \frac{\lambda_X}{\lambda_X + \lambda_Y +\lambda_Z} (X+Y+Z) = \frac{\lambda_X}{\lambda_X + \lambda_Y + \lambda_Z} N$

where we substituted having recognized that the sum of two independent Poisson random variables $Y+Z$ is also a Poisson random variable, say, $W$, whose rate is the sum $\lambda_Y + \lambda_Z$.

Now, consider the joint expectation $\E\biggl[g(X,Y)h(X,Y,Z)\bigg|N\biggr] = \E\biggl[\frac{X}{X+Y+Z} \mathbf{1}_{X+Y+Z \neq 0} \bigg| X+Y+Z\biggr] \ $

We can use the property that $\E[aX] = a\E[X]\ $, and hence,

$\E\biggl[g(X,Y)h(X,Y,Z)\bigg|N\biggr] = \frac{\lambda_X}{\lambda_X+\lambda_Y+\lambda_Z} \mathbf{1}_{X+Y+Z \neq 0} $

Now, consider $\E[X+Y|N]$ $\ldots$

$E[X+Y|X+Y+Z] = \E[W|W+Z] = \frac{\lambda_W}{\lambda_W + \lambda_Z} (W+Z) = \frac{\lambda_X+\lambda_Y}{\lambda_X + \lambda_Y + \lambda_Z} N $

where we made use of another $W$ substitution. Similar to above,

$\E \biggl[h(X,Y,Z) \bigg| N \biggr] = \E\biggl[\frac{X+Y}{X+Y+Z} \mathbf{1}_{X+Y+Z \neq 0} \bigg| X+Y+Z \biggr] = \frac{\lambda_X+\lambda_Y}{\lambda_X+\lambda_Y+\lambda_Z} \mathbf{1}_{X+Y+Z \neq 0}$

Now, one more term remains $\E\biggl[g(X,Y)\bigg| N\biggr]$. We can use the $Law\ of\ Iterated\ Expectations$

$ \E[X|N] = \E\biggl[\E[X|M]\bigg|N\biggr] $

where the value of $N$ is determined by $M$. In our case, $N$ is determined by $Z$ and $X+Y$, both existing on the same probability space. So

$ \E\biggl[g(X,Y) \bigg| N \biggr] = \E \biggl[ \E [g(X,Y)|Z, X+Y)] \bigg| N\biggr] $

But $g(X,Y)$ is independent of $Z$. In which case, focusing on the inner expectation,

$ \E \biggl[g(X,Y)\bigg|Z, X+Y)\biggr] = \E \biggl[g(X,Y)\bigg| X+Y) \biggr] = \E \biggl[ \frac{X}{X+Y} \mathbf{1}_{X+Y \neq 0} \bigg| X+Y \biggr] $

where we have already seen this above.

$ \E \biggl[ \frac{X}{X+Y} \mathbf{1}_{X+Y \neq 0} \bigg| X+Y \biggr] = \frac{\lambda_X}{\lambda_X + \lambda_Y} \mathbf{1}_{X+Y \neq 0} $

If we insert this result into the outer expectation, we have

$ \E \biggl[ g(X,Y) \bigg| N \biggr] = \E \biggl[ \frac{\lambda_X}{\lambda_X + \lambda_Y} \mathbf{1}_{X+Y \neq 0} \bigg| N \biggr] = \frac{\lambda_X}{\lambda_X + \lambda_Y} \mathbf{1}_{X+Y \neq 0}$

where we used the property that the expectation of a constant is equal to the constant ($\E[b] = b$).

Putting the three terms together, we arrive at our solution for the covariance:

$ \Cov(g(X,Y), h(X,Y,Z) | N) = \frac{\lambda_X}{\lambda_X + \lambda_Y + \lambda_Z} - \biggl(\frac{\lambda_X}{\lambda_X + \lambda_Y} \cdot \frac{\lambda_X + \lambda_Y}{\lambda_X + \lambda_Y + \lambda_Z} \biggr) = 0$

So while the functions $g(X,Y)$ and $h(X,Y,Z)$ are clearly dependent, they are not expected to co-vary.

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    @did I think I was able to figure out my question. We should expect zero covariance for a constant $N$ and known $\lambda_x$. But if $N$ varies due to a change in $\lambda_X$ and $\lambda_Y$, we might expect some sort of linear relation between the proportions. But this was not the question I asked! In any event, thanks for your help. I have a related [post](http://math.stackexchange.com/questions/175156/windowed-linear-correlation) building on the above post. Maybe you have some insights?2012-07-25
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Hint: Independent Poisson random variables $U$ and $V$ with respective rates $u$ and $v$ are such that $\mathrm E(U\mid U+V)=\frac{u}{u+v}(U+V)$ hence $\mathrm E\left(\frac{U}{U+V}\,\mathbf 1_{U+V\ne0}\big\vert U+V\right)=\frac{u}{u+v}\,\mathbf 1_{U+V\ne0}$.

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    I provided a partial solution for the covariance, but I am not sure about the '3rd' application of the hint. Any further advice for $\mathrm{E}[g(X,Y)|N]\ $? Thanks again.2012-07-03