Does the "variance effect" hold for any arbitrary distribution?

Question

Let $F(\mu,\sigma)$ be some two parameter distribution, the parameters are mean and std. All rv below are independent.

Let rv $X_0\sim F(0,\sigma)$ and $X_1\sim F(1,\sigma)$

Let $\sigma'\geq \sigma$.

Let rv $X'_0\sim F(0,\sigma')$ and $X'_1\sim F(1,\sigma')$

This means that all these distributions are from the same location-scale family.

Is it true that $P(X_1>X_0)\geq P(X'_1>X_0')$?

It clearly works for normal distributions. I wonder if it also work for any arbitrary distributions. Might be a easy question.

Answer 1

I think what you are trying to ask something equivalent to:

Suppose $S_1,S_2,S_3,S_4$ are independently identically distributed, and for some given $k\ge 1$ and some $c$ you have $T_i=kS_i+c$ (so the standard deviations of the $T_i$s are $k$ times the standard deviations of the $S_i$s). Is it always the case that $$P(S_1 +1\ge S_2)\ge P(T_3+1\ge T_4)\,?$$

To which the answer is yes, since

$$\begin{align} & P(S_1+1\ge S_2) \\ = & P(S_3+1\ge S_4) \\ = & P(kS_3+k\ge kS_4) \\ = & P(kS_3+c+k\ge kS_4+c) \\ = & P(T_3+k\ge T_4) \\ = & P(T_3-T_4\ge -k)\\ \ge & P(T_3-T_4\ge -1) \\ = & P(T_3+1\ge T_4). \end{align}$$

Answer 2

• Instead of making the comparison of variables with a different scale, you can see the equivalent comparison for variables with a different location.

• Obviously a larger location makes the probability larger. Let rv $X_0\sim F(0,\sigma)$, $X_a\sim F(a,\sigma)$ and $X_b\sim F(b,\sigma)$, then for $b>a$ we have

  $$P(X_b>X_0) \geq P(X_a>X_0)$$

• Now to make that shift from comparing variables with different scale to variables with different location, consider a rescaling that preserves the probability but changes the scale and location.

  Let rv $X^\dagger_0\sim F(0,\sigma)$ and $X^\dagger_1\sim F(\sigma/\sigma',\sigma)$.

  The variables $X^\dagger_0$ and $X^\dagger_1$ are just scaled versions of $X'_0$ and $X'_1$ (by a factor $\sigma/\sigma' < 1$) and we have the equality

  $$P(X^\dagger_1>X^\dagger_0) = P(X'_1>X_0')$$

  which means instead of comparing $P(X'_1>X_0')$ with $P(X_1>X_0)$, we can also compare $P(X^\dagger_1>X_0^\dagger)$ with $P(X_1>X_0)$.