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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:
In effect, g(y) is the probability of y occurring and [∫_{−∞}^{y}f(x)dx] is the probability of x being less than y. For convenience let F(y)=∫_{−∞}^{y}f(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
Furthermore,
Now consider ∂R/∂μ at μ=0. We then have
At μ=0 this is
Since lim_{±∞}f(y)=0 and F(y) is everywhere finite the derivative reduces to;
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
where H(y)=∫_{−∞}^{y}h(x/σ)dx.
Now ∂R/∂μ = −(1/σ) ∫_{−∞}^{∞}h'((y−μ/σ)H(y)dy
Letting μ=0 and applying integrationbyparts with U=H(y) and dV=h'(y/σ) gives
Changing the variable of integration from y to z=y/σ then gives
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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