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I'm currently taking a course on game theory and I'm having a hard time understanding why, given a set of pure strategies $S_i\subset \mathbb{R}$, a mixed strategy is defined as the cumulative distribution function (cdf) $F_i:S_i\to [0,1]$.

In the finite case one defines a mixed strategy as a tuple such that its entries reflect the probabilities of playing the corresponding pure strategy. Given that, I would have expected $F_i$ to be the probability density function (pdf) in the non-finite case rather than the cdf.

Is this just a convention and it is equally fine to define a mixed strategy as the pdf? If not, can anyone explain to me why cdf is the right way to go? If it is just a convention, why do people chose cdf over pdf?

Cheers

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    I am not familiar with game-theory, but a distribution is completely determined by its **unique** and **always existing** cdf. A pdf might not exist and if it does then in most cases it is not unique. These could be arguments to hold on to cdf's here. – 2017-02-05
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    Thanks @drhab ! Assuming what I wrote in the first two paragraphs is correct, that would be a satisfying explanation. – 2017-02-05
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    After having thought about it a little more I think that in game theory cdf's always have to be differentiable, since one can only define expected payoff using pdf's. So maybe existence is not the real reason behind it and uniqueness alone is a little thin to this end. – 2017-02-05
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    @FloodLuszt drhab is correct. A PDF will always have a corresponding CDF but a given CDF may not always have an equivalent PDF. CDFs do not always have to be differentiable when doing game theory. We just typically assume mixed strategies which have a PDF for analytical tractability (in the same way that one might assume consumers have differentiable utility functions). – 2017-02-05
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    @TheoreticalEconomist great thanks. One more question though: How does one define expected payoffs in the case of non-differentiable CDFs? – 2017-02-06
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    Nevermind you can always define it as $E(u)=\intop u(x) \;\mathrm{d}F(x)$. Thanks everyone. – 2017-02-06

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