# maximizing probability that one variable is greater than another

#### Vova

##### New Member
hi,
There is a random variable X with a known distribution along its support interval, say, [0,a].

Need to find a distribution function of another independent random variable Y with given expected value E(Y) and distributing along the same interval [0,a] so that P(Y>X) is maximal.

here http://math.stackexchange.com/quest...ability-that-one-random-variable-is-less-than is explained how to calculate the probability that one variable is greater than another, but how to maximize it?

In my particular example X has a mass point at zero and uniformly distributes on (0,a]. My solution candidate Y has a mass at a and uniformly distributes on [0,a), but I'm unable to show that Y indeed maximizes P(Y>X). Also, as I've said before, I'm looking for a general method to determine Y.

#### BGM

##### TS Contributor
General formulation:

Given a random variable $$X$$ with CDF $$F_X$$, find a random variable $$Y$$ (i.e. equivalent to find its CDF $$F_Y$$) to

Maximize
$$\Pr\{Y > X\} = \int_{-\infty}^{+\infty} \int_{-\infty}^y dF_X(x)dF_Y(y)$$

subject to

$$\int_{-\infty}^{+\infty} ydF_Y(y) = \mu$$

where $$\mu < +\infty$$ is a given constant.

This, in a general Mathematical context should be related something like functional optimization, which I cannot go into details here as I do not know much either.

For your particular example, I can try to give a heuristic guess for it.

Let the point mass at 0 for $$X$$ is $$p$$.

Lets test some trivial cases first:

If you set $$Y = \mu$$ to be a constant, then

$$\Pr\{Y > X\} = \Pr\{X < \mu\} = p + (1 - p)\frac {\mu} {a}$$

If you set $$Y$$ to be a discrete random variable with 2 support points $$\{y_1, y_2\}$$ such that $$0 < y_1 < \mu < y_2 < a$$ and

$$p_Yy_1 + (1 - p_Y)y_2 = \mu$$

then

$$\Pr\{Y > X\} = \Pr\{X < y_1\}p_Y + \Pr\{X < y_2\}(1 - p_Y)$$

$$= \left[p + (1 - p)\frac {y_1} {a}\right]p_Y + \left[p + (1 - p)\frac {y_2} {a}\right](1 - p_Y)$$

$$= p + (1 - p)\frac {\mu} {a}$$

which is the same. Inductively, for any discrete random variable without support on $$0$$, the probability is the same. And it is easy to see that for any discrete random variable with support $$0$$ it must be worse.

For $$Y$$ to be a purely continuous random variable, with

$$\int_0^a yf_Y(y)dy = \mu$$

Then

$$\Pr\{Y > X\}$$

$$= \int_0^a \Pr\{X < y\}f_Y(y)dy$$

$$= \int_0^a \left[p + (1 - p)\frac {y} {a}\right] f_Y(y)dy$$

$$= p + (1 - p) \frac {\mu} {a}$$

Again it has the same probability.

For mixed case can work out need to work out tomorrow.