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The Ratio of Winners for a Stochastic Competition

Let X and Y be two continuous stochastic variables with probability distributions f(x) and g(y), respectively. If a random selection of x and y has x less than y then X is deemed a winner. Since x and y have continuous values no special case will be specified for the case of x=y as the likelihood of x and y being equal is zero. The probability of X winning is denoted as R. The value of R is given by:

R = ∫−∞g(y)[∫−∞yf(x)dx]dy

In effect, g(y) is the probability of y occurring and [∫−∞yf(x)dx] is the probability of x being less than y. For convenience let F(y)=∫−∞yf(x)dx.

Now consider the case in which g(y)=f(y−μ); i.e., the shapes of the distributions are the same, one is just shifted with respect to the other. If μ=0 then from symmetry we know that R=0.5. This also follows analytically because f(y)=F'(y) so

−∞f(y)F(y)dy = ½[F²(y)|−∞] = ½[1−0] = ½

Furthermore,

limμ→+∞ R = 1.0
and
limμ→−∞ R = 0.0

Now consider ∂R/∂μ at μ=0. We then have

R = ∫−∞f(y−μ)F(y)dy
so
∂R/∂μ = −∫−∞f'(y−μ)F(y)dy

At μ=0 this is

(∂R/∂μ)μ=0 = −∫−∞f'(y)F(y)dy
which upon applying integration-by-parts
and noting that F'(y)=f(y) gives
(∂R/∂μ)μ=0 = −[f(y)F(y)|−∞ − ∫−∞+∞f²(y)dy]

Since lim±∞f(y)=0 and F(y) is everywhere finite the derivative reduces to;

(∂R/∂μ)μ=0 = ∫−∞+∞f²(y)dy

For the normal distribution f(y)=exp(−(½(y−μ/σ)²) the above integral evaluates to 1/(2√πσ)=0.2821/σ. For the negative exponential distribution f(y)=exp(−y/α)/α the above integral is equal to 1/(2α). In both of these cases the sensitivity of the ratio to changes in the mean is inversely proportional to the degree of dispersion of the distribution.

Now going back to the original formulation consider the case in which f(x)=h(x/σ) and g(y)=h((y−μ)/σ). The win ratio for X is then

R = ∫−∞h((y−μ)/σ)H(y)dy

where H(y)=∫−∞yh(x/σ)dx.

Now ∂R/∂μ = −(1/σ) ∫−∞h'((y−μ/σ)H(y)dy

Letting μ=0 and applying integration-by-parts with U=H(y) and dV=h'(y/σ) gives

(∂R/∂μ)μ=0 = ∫−∞h²(y/σ)dy

Changing the variable of integration from y to z=y/σ then gives

(∂R/∂μ)μ=0 = (1/σ) ∫−∞h²(z)dz

It is notable that the derivative (∂R/∂μ) is inversely proportional to the degree of dispersion of the distribution.

(To be continued.)


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