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Discrete versus continuous operational calculus in antagonistic stochastic games

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Abstract

We consider a class of antagonistic stochastic games in real time between two players A and B, formalized by two marked point processes. The players attack each other at random times with random impacts. Either player can sustain casualties up to a fixed threshold. A player is defeated when its underlying threshold is crossed. Upon that time (referred to as the first passage time), the game is over. We introduce a joint functional of the first passage time, along with the status of each player upon this time. The latter are the cumulative magnitudes of casualties to each player upon the end of the game, obtained in an analytically tractable form. We then use discrete and continuous operational calculus for the transform inversion. We demonstrate in a special case that the discrete operational calculus is more efficient, allowing us to avoid numerical inversion. It leads to explicit formulas for the joint distribution of associated random variables (first passage time and the status of cumulative casualties to the players upon the end of the game).

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Acknowledgements

The authors are indebted to anonymous referees who made numerous very valuable suggestions that hugely improved the paper.

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Correspondence to Jewgeni H. Dshalalow.

Appendix

Appendix

Proof of Theorem 4

In the sequel, for brevity we denote \(\eta =\gamma \left( \theta \right) =\frac{\gamma }{\gamma +\theta }\). From (3.1) and (3.2) we have

$$\begin{aligned} \frac{1}{1-\gamma (ux,vy,\theta )}=\frac{1}{1-\frac{a}{1-\alpha ux}\frac{b}{1-\beta vy}\eta }. \end{aligned}$$

After a straightforward algebra we get

$$\begin{aligned} \frac{1}{1-\gamma (ux,vy,\theta )}=1+\left( \frac{1-\beta vy}{ab\eta }-1\right) ^{-1}\frac{1}{1-\xi x} \end{aligned}$$

where

(A.0)
(A.1)

Continuing with the expression in braces we arrive at

$$\begin{aligned} \left( \frac{1-\beta vy}{ab\eta }-1\right) ^{-1}\frac{1}{1-\xi }= & {} \frac{ab\eta }{1-ab\eta -\alpha u-\left( \beta v-\alpha \beta uv\right) y}\nonumber \\= & {} \frac{ab\eta }{1-ab\eta -\alpha u}\frac{1}{1-\frac{\beta v-\alpha \beta uv}{1-ab\eta -\alpha u}y}. \end{aligned}$$
(A.2)

Furthermore,

$$\begin{aligned} \xi ^M= & {} \left( \frac{\alpha u-\alpha \beta uvy}{1-\beta vy-ab\eta }\right) ^M=\left( \alpha u+\frac{ab\eta \alpha u}{1-\beta vy-ab\eta }\right) ^M\\= & {} \left( \alpha u\right) ^M\sum _{k=0}^M\left( {\begin{array}{c}M\\ k\end{array}}\right) \left( ab\eta \right) ^k\frac{1}{\left( 1-ab \eta -\beta vy\right) ^k} \end{aligned}$$

implying that

$$\begin{aligned}&\left( \frac{1-\beta vy}{ab\eta }-1\right) ^{-1}\frac{\xi ^M}{1-\xi }\nonumber \\&\quad =\frac{\left( \alpha u\right) ^M\sum _{k=0}^M\left( {\begin{array}{c}M\\ k\end{array}}\right) \left( ab\eta \right) ^{k+1}}{\left( 1-ab\eta - \alpha u\right) \left( 1-ab\eta \right) ^k}\frac{1}{\left( 1-\frac{\beta v-\alpha \beta uv}{1-ab\eta -\alpha u}y\right) }\frac{1}{\left( 1-\frac{\beta v}{1-ab\eta }y\right) ^k}. \end{aligned}$$
(A.3)

Substituting (A.2) and (A.3) in (A.1) and applying Theorem 3(iv) and (vi), after some algebra, we get

$$\begin{aligned}&\mathcal {D}_{xy}^{M-1,N-1}\left\{ \frac{1}{1-\gamma (ux,vy,\theta )} \right\} \\&\quad =1+\frac{ab\eta }{1-ab\eta -\alpha u}\mathcal {D}_y^{N-1}\left\{ \frac{1}{1-\frac{\beta v-\alpha \beta uv}{1-ab\eta -\alpha u}y}\right\} \\&\qquad -\sum _{k=0}^M\left( {\begin{array}{c}M\\ k\end{array}}\right) \left( ab\eta \right) ^{k+1}\frac{\left( \alpha u\right) ^M}{\left( 1-ab\eta -\alpha u\right) \left( 1-ab\eta \right) ^k}\mathcal {D}_y^{N-1}\\&\qquad \left\{ \frac{1}{\left( 1- \frac{\beta v-\alpha \beta uv}{1-ab\eta -\alpha u}y\right) }\frac{1}{\left( 1-\frac{\beta v}{1-ab\eta }y\right) ^k}\right\} \\&\quad =1+\left[ 1-\left( \alpha u\right) ^M\right] \frac{ab\eta }{1-ab\eta -\alpha u-\beta v+\alpha \beta uv}\left[ 1-\left( \frac{\beta v-\alpha \beta uv}{1-ab\eta -\alpha u}\right) ^N\right] \\&\qquad -\left\{ \left( \alpha u\right) ^M\sum _{k=1}^M\left( {\begin{array}{c}M\\ k\end{array}}\right) \left( ab\eta \right) ^{k+1}\frac{1}{ \left( 1-ab\eta \right) ^k}\frac{1}{1-ab\eta -\alpha u-\left( \beta v-\alpha \beta uv\right) }\right. \\&\qquad \left. \times \sum _{j=0}^{N-1}\left( {\begin{array}{c}k+j-1\\ j\end{array}}\right) \left( \frac{\beta v}{1-ab\eta }\right) ^j\left[ 1-\left( \frac{\beta v-\alpha \beta uv}{1-ab\eta -\alpha u}\right) ^{N-j}\right] \right\} . \end{aligned}$$

Using formula (1.7) of Theorem 1 and recalling that \(\eta =\frac{\gamma }{\gamma +\theta }\) we arrive at formula (3.3) and herewith complete the proof. \(\square \)

Remark 1

In the proof of Theorem 4, we tacitly assumed that \(\xi =\frac{\alpha u\left( 1-\beta vy\right) }{1-\beta vy-ab\eta }\ne 1\) of (A.0). It seems as if we missed to consider \(\xi =1\) as a distinct case. First we note that formally, by Theorem 3 (iv), \(\mathcal {D}^{M-1}_x\frac{1}{1-\xi x}=M\) and thus this version of (A.1) would read

$$\begin{aligned}&\mathcal {D}_{xy}^{M-1,\,N-1}\left\{ \frac{1}{1-\gamma (ux,vy,\, \theta )}\right\} \\&\quad =\mathcal {D}_y^{N-1}\circ \mathcal {D}_x^{M-1}\left\{ 1+\left( \frac{1-\beta vy}{ab\eta }-1\right) ^{-1}\frac{1}{1-\xi x}\right\} \\&\quad =\mathcal {D}_y^{N-1}\left\{ 1+\left( \frac{1-\beta vy}{ab\eta }-1\right) ^{-1}M\right\} . \end{aligned}$$

However, under this assumption, \(y,\alpha ,a,\theta ,u,v\) end up being interdependent which is unacceptable. We therefore see that \(\xi \) cannot be equal 1 or any other constant. For example, with the choice of \(u=v=1,\theta =0,\) not only will y and constants \(\alpha ,a,\gamma \) become dependent, but it eventually leads to forcing y to be a constant, which is absurd. \(\square \)

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Dshalalow, J.H., Iwezulu, K. Discrete versus continuous operational calculus in antagonistic stochastic games. São Paulo J. Math. Sci. 11, 471–489 (2017). https://doi.org/10.1007/s40863-017-0073-9

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