Appendix: A
The analytic expression for
$$\mathbb E\left[ \tilde P_{A0}\tilde P_{B0}\right] =\sum\limits_{j = 1}^{+\infty}\sum\limits_{p = 1}^{+\infty } \frac{\mathbb E\left[ \tilde d_{Aj}\tilde d_{Bp}\right]}{(1 + k_{A})^{j}(1 + k_{B})^{p}} $$
is obtained recalling (12). It results that
$$ \mathbb E\left[ \tilde P_{A0}\tilde P_{B0}\right] = d_{A0}d_{B0}\sum\limits_{p = 1}^{+\infty} \left( \underset{\text{\#1}}{\underbrace{\sum\limits_{j = 1}^{p}\frac{(1 + G_{A})^{j}(1 + \bar g_{B})^{p}}{(1 + k_{A})^{j}(1 + k_{B})^{p}}}} + \underset{\text{\#2}}{\underbrace{\sum\limits_{j = p + 1}^{+\infty}\frac{(1 + \bar g_{A})^{j}(1 + G_{B})^{p}}{(1 + k_{A})^{j}(1 + k_{B})^{p}}}}\right). $$
(19)
For ease of notation, let, for i = A,B,
$$\gamma_{\bar g_{i}} = \frac{1 + \bar g_{i}}{1 + k_{i}} \hspace{1cm} \text{and} \hspace{1cm} \gamma_{G_{i}} = \frac{1 + G_{i}}{1 + k_{i}}. $$
Sum #1 becomes
$$\gamma_{\bar g_{B}}^{p}\sum\limits_{j = 1}^{p}\gamma_{G_{A}}^{j} =\frac{\gamma_{G_{A}}}{1 - \gamma_{G_{A}}}\left( 1 - \gamma_{G_{A}}^{p}\right)\gamma_{\bar g_{B}}^{p}, $$
while sum #2, that converges to a positive value as soon as \(\bar g_{A} < k_{A}\), reduces to
$$\gamma_{G_{B}}^{p}\sum\limits_{j = p + 1}^{+\infty}\gamma_{\bar g_{A}}^{j} =\frac{\gamma_{\bar g_{A}}}{ 1-\gamma_{\bar g_{A}}}\left( \gamma_{G_{B}}\gamma_{\bar g_{A}}\right)^{p}. $$
Observing that
$$\gamma_{G_{B}}\gamma_{\bar g_{A}} =\frac{\left( 1 + \bar g_{A}\right)\left( 1 + \bar g_{B}\right) + \text{Cov}\left[\tilde g_{A},\tilde g_{B}\right]}{\left( 1 + k_{A}\right)\left( 1 + k_{B}\right)} =\gamma_{G_{A}}\gamma_{\bar g_{B}}, $$
sum (19) results being
$$\begin{array}{@{}rcl@{}} \mathbb E \left[ \tilde P_{A0}\tilde P_{B0}\right] & =& d_{A0}d_{B0}\sum\limits_{p = 1}^{+\infty}\left( \frac{\gamma_{G_{A}}}{1 - \gamma_{G_{A}}}\gamma_{\bar g_{B}}^{p} - \frac{\gamma_{G_{A}}}{1 - \gamma_{G_{A}}} \left( \gamma_{G_{A}}\gamma_{\bar g_{B}}\right)^{p} + \frac{\gamma_{\bar g_{A}}}{1 - \gamma_{\bar g_{A}}}\left( \gamma_{G_{B}}\gamma_{\bar g_{A}}\right)^{p}\right)\\ & =& d_{A0}d_{B0}\left( \frac{\gamma_{G_{A}}}{1 - \gamma_{G_{A}}}\sum\limits_{p = 1}^{+\infty }\gamma_{\bar g_{B}}^{p} +\left( \frac{\gamma_{\bar g_{A}}}{1 - \gamma_{\bar g_{A}}} - \frac{\gamma_{G_{A}}}{1 - \gamma_{G_{A}}}\right)\sum\limits_{p = 1}^{+\infty }\left( \gamma_{G_{B}}\gamma_{\bar g_{A}}\right)^{p}\right). \end{array} $$
Now, \({\sum }_{p = 1}^{+\infty }\gamma _{\bar g_{B}}^{p}\) converges to
$$\frac{\gamma_{\bar g_{B}}}{1 - \gamma_{\bar g_{B}}} = \frac{1 + \bar g_{B}}{k_{B} - \bar g_{B}} > 0 $$
if \(\bar g_{B} < k_{B}\), whereas \({\sum }_{p = 1}^{+\infty }\left (\gamma _{G_{B}}\gamma _{\bar g_{A}}\right )^{p}\) converges to
$$\frac{\gamma_{G_{A}}\gamma_{\bar g_{B}}}{1 - \gamma_{G_{A}}\gamma_{\bar g_{B}}} =\frac{\left( 1 + \bar g_{A}\right)\left( 1 + \bar g_{B}\right) + \text{Cov}\left[\tilde g_{A},\tilde g_{B}\right]} {\left( 1 + k_{A}\right) \left( 1 + k_{B}\right) - \left( 1 + \bar g_{A}\right)\left( 1 + \bar g_{B}\right) - \text{Cov}\left[\tilde g_{A},\tilde g_{B}\right]} $$
if
$$\vert\left( 1 + \bar g_{A}\right)\left( 1 + \bar g_{B}\right)+ \text{Cov}\left[ \tilde g_{A},\tilde g_{B}\right]\vert < \left( 1 + k_{A}\right)\left( 1 + k_{B}\right). $$
Moreover, observe that
$$\bar P_{m0} = \frac{d_{m0}\gamma_{\bar g_{m}}}{1 - \gamma_{\bar g_{m}}},\hspace{1cm} m = A,B. $$
Covariance between \(\tilde P_{A0}\) and \(\tilde P_{B0}\) is, then,
$$\begin{array}{@{}rcl@{}} && \text{Cov}\left[ \tilde P_{A0}\tilde P_{B0}\right]\\ & =& d_{A0}d_{B0}\left( \frac{\gamma_{G_{A}}}{1 - \gamma_{G_{A}}}\frac{\gamma_{\bar g_{B}}}{1 - \gamma_{\bar g_{B}}} +\left( \frac{\gamma_{\bar g_{A}}}{1 - \gamma_{\bar g_{A}}} - \frac{\gamma_{G_{A}}}{1 - \gamma_{G_{A}}}\right)\frac{\gamma_{G_{A}}\gamma_{\bar g_{B}}}{1 - \gamma_{G_{A}}\gamma_{\bar g_{B}}} - \frac{\gamma_{\bar g_{A}}}{1 - \gamma_{\bar g_{A}}}\frac{\gamma_{\bar g_{B}}}{1 - \gamma_{\bar g_{B}}}\right)\\ & =& d_{A0}d_{B0}\left( \frac{\gamma_{G_{A}}}{1 - \gamma_{G_{A}}} - \frac{\gamma_{\bar g_{A}}}{1 - \gamma_{\bar g_{A}}}\right) \left( \frac{\gamma_{\bar g_{B}}}{1 - \gamma_{\bar g_{B}}} -\frac{\gamma_{G_{A}}\gamma_{\bar g_{B}}}{1 - \gamma_{G_{A}}\gamma_{\bar g_{B}}}\right)\\ & =& d_{A0}d_{B0} \left( \frac{\gamma_{G_{A}} - \gamma_{\bar g_{A}}}{\left( 1 - \gamma_{G_{A}}\right)\left( 1 - \gamma_{\bar g_{A}}\right)}\right) \left( \frac{\gamma_{\bar g_{B}}\left( 1 - \gamma_{G_{A}}\right)}{\left( 1 - \gamma_{\bar g_{B}}\right)\left( 1 - \gamma_{G_{A}}\gamma_{\bar g_{B}}\right)}\right)\\ & =& \left( \frac{d_{A0}\gamma_{\bar g_{A}}}{1 - \gamma_{\bar g_{A}}}\right) \left( \frac{d_{B0}\gamma_{\bar g_{B}}}{1 - \gamma_{\bar g_{B}}}\right) \left( \frac{\gamma_{G_{A}} - \gamma_{\bar g_{A}}}{\gamma_{\bar g_{A}}\left( 1 - \gamma_{G_{A}}\gamma_{\bar g_{B}}\right)}\right)= \frac{\bar P_{A0}\bar P_{B0}\left( \gamma_{G_{A}} - \gamma_{\bar g_{A}}\right)}{\gamma_{\bar g_{A}}\left( 1 - \gamma_{G_{A}}\gamma_{\bar g_{B}}\right)}. \end{array} $$
(20)
Now, recall the definition of \(\gamma _{G_{A}}\), \(\gamma _{\bar g_{A}}\), \(\gamma _{\bar g_{B}}\), and GA. Then,
$$\gamma_{G_{A}} - \gamma_{\bar g_{A}} =\frac{\text{Cov}\left[\tilde g_{A},\tilde g_{B}\right]}{(1 + k_{A})(1 + \bar g_{B})}, $$
and
$$\gamma_{\bar g_{A}}\left( 1 - \gamma_{G_{A}}\gamma_{\bar g_{B}}\right) =\frac{\left( 1 + \bar g_{A}\right)\left( (1 + k_{A})(1 + k_{B}) - (1 + \bar g_{A})(1 + \bar g_{B}) - \text{Cov}\left[\tilde g_{A},\tilde g_{B}\right]\right)}{(1 + k_{A})^{2}(1 + k_{B})} $$
Substituting these expressions in (20),
$$\begin{array}{@{}rcl@{}} && \text{Cov}\left[ \tilde P_{A0}\tilde P_{B0}\right]\\ & =& \bar P_{A0}\bar P_{B0}\times \frac{\text{Cov}\left[\tilde g_{A},\tilde g_{B}\right]}{(1 + k_{A})(1 + \bar g_{B})} \times\frac{(1 + k_{A})^{2}(1 + k_{B})}{(1 + \bar g_{A})\left( (1 + k_{A})(1 + k_{B}) - (1 + \bar g_{A})(1 + \bar g_{B}) - \text{Cov}\left[\tilde g_{A},\tilde g_{B}\right]\right)}\\ & =& \bar P_{A0}\bar P_{B0}\times \frac{(1 + k_{A})(1 + k_{B})}{(1 + \bar g_{A})(1 + \bar g_{B})} \times\frac{\text{Cov}\left[\tilde g_{A},\tilde g_{B}\right]}{(1 + k_{A})(1 + k_{B}) - (1 + \bar g_{A})(1 + \bar g_{B}) - \text{Cov}\left[\tilde g_{A},\tilde g_{B}\right]}. \end{array} $$
This proves the claimed formula.
Appendix: B
The analytic expression for
$$\mathbb E\left[ \tilde P_{At}\tilde P_{Bt}\right] =\sum\limits_{j = t + 1}^{+\infty}\sum\limits_{p = t + 1}^{+\infty } \frac{\mathbb E\left[ \tilde d_{Aj}\tilde d_{Bp}\right]}{(1 + k_{A})^{j - t}(1 + k_{B})^{p - t}} $$
is obtained analogously to that of \(\mathbb E\left [ \tilde P_{A0}\tilde P_{B0}\right ]\) by means of the substitutions u := j − t and v := p − t, and recalling that
$$\mathbb E\left[ \tilde d_{A,u + t}\tilde d_{B,v + t}\right] =\left\{\begin{array}{ll} d_{A0}d_{B0}\left( 1 + G_{A}\right)^{u + t}\left( 1 + \bar g_{B}\right)^{v + t} u \leq v \\ d_{A0}d_{B0}\left( 1 + G_{B}\right)^{v + t}\left( 1 + \bar g_{A}\right)^{u + t} u >v. \end{array}\right. $$
Therefore
$$ \mathbb E\left[ \tilde P_{At}\tilde P_{Bt}\right] = d_{A0}d_{B0}\sum\limits_{v = 1}^{+\infty} \left( \underset{\text{\#1}}{\underbrace{\sum\limits_{u = 1}^{v}\frac{(1 + G_{A})^{u + t}(1 + \bar g_{B})^{v + t}}{(1 + k_{A})^{u}(1 + k_{B})^{v}}}} + \underset{\text{\#2}}{\underbrace{\sum\limits_{u = v + 1}^{+\infty}\frac{(1 + \bar g_{A})^{u + t}(1 + G_{B})^{v + t}}{(1 + k_{A})^{u}(1 + k_{B})^{v}}}}\right). $$
(21)
Under the assumptions of Proposition 1, sum #1 becomes
$$(1 + G_{A})^{t}(1 + \bar g_{B})^{t}\gamma_{\bar g_{B}}^{v}\sum\limits_{u = 1}^{v}\gamma_{G_{A}}^{u} =(1 + G_{A})^{t}(1 + \bar g_{B})^{t}\frac{\gamma_{G_{A}}}{1 - \gamma_{G_{A}}}\left( 1 - \gamma_{G_{A}}^{v}\right)\gamma_{\bar g_{B}}^{v}, $$
while sum #2 reduces to
$$(1 + G_{B})^{t}(1 + \bar g_{A})^{t}\gamma_{G_{B}}^{v}\sum\limits_{u = v + 1}^{+\infty}\gamma_{\bar g_{A}}^{u} =(1 + G_{B})^{t}(1 + \bar g_{A})^{t}\frac{\gamma_{\bar g_{A}}}{ 1-\gamma_{\bar g_{A}}}\left( \gamma_{G_{B}}\gamma_{\bar g_{A}}\right)^{v}. $$
Recalling that
$$(1 + G_{A})(1 + \bar g_{B}) =(1 + G_{B})(1 + \bar g_{A}) =(1 + \bar g_{A})(1 + \bar g_{B}) + \text{Cov}\left[\tilde g_{A},\tilde g_{B}\right], $$
and that \(\gamma _{G_{B}}\gamma _{\bar g_{A}} = \gamma _{G_{A}}\gamma _{\bar g_{B}}\), sum (21) results being
$$\begin{array}{@{}rcl@{}} &&\mathbb E \left[ \tilde P_{At}\tilde P_{Bt}\right]\\ & = &f(t) \sum\limits_{v = 1}^{+\infty}\left( \frac{\gamma_{G_{A}}}{1 - \gamma_{G_{A}}}\gamma_{\bar g_{B}}^{v} - \frac{\gamma_{G_{A}}}{1 - \gamma_{G_{A}}} \left( \gamma_{G_{A}}\gamma_{\bar g_{B}}\right)^{v} + \frac{\gamma_{\bar g_{A}}}{1 - \gamma_{\bar g_{A}}}\left( \gamma_{G_{B}}\gamma_{\bar g_{A}}\right)^{v}\right)\\ & =& f(t) \left( \frac{\gamma_{G_{A}}}{1 - \gamma_{G_{A}}}\sum\limits_{v = 1}^{+\infty }\gamma_{\bar g_{B}}^{v} +\left( \frac{\gamma_{\bar g_{A}}}{1 - \gamma_{\bar g_{A}}} - \frac{\gamma_{G_{A}}}{1 - \gamma_{G_{A}}}\right){\sum}_{v = 1}^{+\infty }\left( \gamma_{G_{A}}\gamma_{\bar g_{B}}\right)^{v}\right)\\ & =& f(t) \left( \frac{\gamma_{G_{A}}}{1 - \gamma_{G_{A}}}\frac{\gamma_{\bar g_{B}}}{1 - \gamma_{\bar g_{B}}} +\left( \frac{\gamma_{\bar g_{A}}}{1 - \gamma_{\bar g_{A}}} - \frac{\gamma_{G_{A}}}{1 - \gamma_{G_{A}}}\right)\frac{\gamma_{G_{A}}\gamma_{\bar g_{B}}}{1 - \gamma_{G_{A}}\gamma_{\bar g_{B}}}\right), \end{array} $$
where
$$f(t) = d_{A0}d_{B0}\left( (1 + \bar g_{A})(1 + \bar g_{B}) + \text{Cov}\left[\tilde g_{A},\tilde g_{B}\right]\right)^{t}. $$
Observing that
$$f(t) = \bar P_{A0}\bar P_{B0}\left( (1 + \bar g_{A})(1 + \bar g_{B}) + \text{Cov}\left[\tilde g_{A},\tilde g_{B}\right]\right)^{t} \frac{1 - \gamma_{\bar g_{A}}}{\gamma_{\bar g_{A}}}\frac{1 - \gamma_{\bar g_{B}}}{\gamma_{\bar g_{B}}}, $$
one can write
$$\mathbb E \left[ \tilde P_{At}\tilde P_{Bt}\right] = g(t) \left( \frac{\gamma_{G_{A}}}{1 - \gamma_{G_{A}}}\frac{1 - \gamma_{\bar g_{A}}}{\gamma_{\bar g_{A}}} + \left( 1 - \frac{\gamma_{G_{A}}\left( 1 - \gamma_{\bar g_{A}}\right)}{\gamma_{\bar g_{A}}\left( 1 - \gamma_{G_{A}}\right)}\right) \frac{\gamma_{G_{A}}\left( 1 - \gamma_{\bar g_{B}}\right)}{1 - \gamma_{G_{A}}\gamma_{\bar g_{B}}}\right), $$
where
$$g(t) = \bar P_{A0}\bar P_{B0}\left( (1 + \bar g_{A})(1 + \bar g_{B}) + \text{Cov}\left[\tilde g_{A},\tilde g_{B}\right]\right)^{t}. $$
Trivial algebra proves that
$$\frac{\gamma_{G_{A}}}{1 - \gamma_{G_{A}}}\frac{1 - \gamma_{\bar g_{A}}}{\gamma_{\bar g_{A}}} +\left( 1 - \frac{\gamma_{G_{A}}\left( 1 - \gamma_{\bar g_{A}}\right)}{\gamma_{\bar g_{A}}\left( 1 - \gamma_{G_{A}}\right)}\right) \frac{\gamma_{G_{A}}\left( 1 - \gamma_{\bar g_{B}}\right)}{1 - \gamma_{G_{A}}\gamma_{\bar g_{B}}} =\frac{\gamma_{G_{A}}\left( 1 - \gamma_{\bar g_{A}}\gamma_{\bar g_{B}}\right)}{\gamma_{\bar g_{A}}\left( 1 - \gamma_{G_{A}}\gamma_{\bar g_{B}}\right)}, $$
so that
$$\mathbb E \left[ \tilde P_{At}\tilde P_{Bt}\right] =\bar P_{A0}\bar P_{B0}\frac{\gamma_{G_{A}}\left( 1 - \gamma_{\bar g_{A}}\gamma_{\bar g_{B}}\right)}{\gamma_{\bar g_{A}}\left( 1 - \gamma_{G_{A}}\gamma_{\bar g_{B}}\right)} \left( (1 + \bar g_{A})(1 + \bar g_{B}) + \text{Cov}\left[\tilde g_{A},\tilde g_{B}\right]\right)^{t}. $$
Moreover, for m = A,B we have
$$\bar P_{mt} =\mathbb E\left[\sum\limits_{j = t + 1}^{+\infty}\frac{\tilde d_{mj}}{(1 + k_{m})^{j - t}}\right] =\mathbb E\left[\sum\limits_{u = 1}^{+ \infty}\frac{\tilde d_{m,u + t}}{\left( 1 + k_{m}\right)^{u}}\right] =\bar P_{m0}\left( 1 + \bar g_{m}\right)^{t}, $$
and covariance between \(\tilde P_{At}\) and \(\tilde P_{Bt}\) is
$$\begin{array}{@{}rcl@{}} && \text{Cov}\left[ \tilde P_{At}\tilde P_{Bt}\right]\\ & =& \bar P_{A0}\bar P_{B0}\frac{\gamma_{G_{A}}\left( 1 - \gamma_{\bar g_{A}}\gamma_{\bar g_{B}}\right)}{\gamma_{\bar g_{A}}\left( 1 - \gamma_{G_{A}}\gamma_{\bar g_{B}}\right)}\left( (1 + \bar g_{A})(1 + \bar g_{B}) + \text{Cov}\left[\tilde g_{A},\tilde g_{B}\right]\right)^{t} -\bar P_{A0}\bar P_{B0}\left( 1 + \bar g_{A}\right)^{t}\left( 1 + \bar g_{B}\right)^{t}\\ & =& \bar P_{A0}\bar P_{B0}\left( 1 + \bar g_{A}\right)^{t}\left( 1 + \bar g_{B}\right)^{t} \left( \frac{\gamma_{G_{A}}\left( 1 - \gamma_{\bar g_{A}}\gamma_{\bar g_{B}}\right)}{\gamma_{\bar g_{A}}\left( 1 - \gamma_{G_{A}}\gamma_{\bar g_{B}}\right)} \left( 1 + \frac{\text{Cov}\left[\tilde g_{A},\tilde g_{B}\right]}{(1 + \bar g_{A})(1 + \bar g_{B})}\right)^{t} -1\right). \end{array} $$
Recalling the definitions of \(\gamma _{G_{A}}\), \(\gamma _{\bar g_{A}}\), \(\gamma _{\bar g_{B}}\), and GA,
$$\frac{\gamma_{G_{A}}\left( 1 - \gamma_{\bar g_{A}}\gamma_{\bar g_{B}}\right)}{\gamma_{\bar g_{A}}\left( 1 - \gamma_{G_{A}}\gamma_{\bar g_{B}}\right)} =\left( 1 + \frac{\text{Cov}\left[\tilde g_{A},\tilde g_{B}\right]}{(1 + \bar g_{A})(1 + \bar g_{B})}\right) \frac{(1 + k_{A})(1 + k_{B}) - (1 + \bar g_{A})(1 + \bar g_{B})}{(1 + k_{A})(1 + k_{B}) - (1 + \bar g_{A})(1 + \bar g_{B}) - \text{Cov}\left[\tilde g_{A},\tilde g_{B}\right]}, $$
and the claimed formula holds.
