# When does one normal distribution dominate another?

Continuing the study of the dominance relationship I defined previously (and only I care about) I thought to ask the question “When does one normal random variable dominate another?”. The answer is very easy to work out, but I found it surprising until I actually did the maths.

Theorem: Let $$A \sim \mathrm{Norm}(\mu_1, \sigma_1^2)$$, $$B \sim \mathrm{Norm}(\mu_2, \sigma_2^2)$$. Then $$A \preceq B$$ iff $$\mu_1 \leq \mu_2$$ and $$\sigma_1 = \sigma_2$$.

Note the equality in the second part: Given two normal distributions with different variance, neither will dominate th e other.

Proof:

Let $$G(t) = P(Z \geq t)$$ where $$Z \sim \mathrm{Norm}(0,1)$$. Then $$P(A \geq t) = G(\frac{t-\mu_1}{\sigma_1})$$, $$P(B \geq t) = G(\frac{t-\mu_2}{\sigma_2})$$. $$G$$ is strictly decreasing, so $$P(A \geq t) \leq P(B \geq t)$$ iff $$\frac{t-\mu_1}{\sigma_1} \geq \frac{t-\mu_2}{\sigma_2}$$ iff $$\left(\frac{1}{\sigma_1} – \frac{1}{\sigma_2}\right) t \geq \frac{\mu_1}{\sigma_1} – \frac{\mu_2}{\sigma_2}$$.

Because the left hand side is linear in $$t$$, this can only be satisfied for all $$t$$ if the coefficient is 0. i.e. if $$\sigma_1 = \sigma_2$$. In this case it is satisfied iff $$0 \geq \frac{\mu_1}{\sigma_1} – \frac{\mu_2}{\sigma_2} = \frac{\mu_1 – \mu_2}{\sigma_1}$$ i.e. iff $$\mu_1 \leq \mu_2$$. Running this backwards, if $$\mu_1 \leq \mu_2$$ and $$\sigma_1 = \sigma_2$$ then this inequality is always satisfies and thus $$A \preceq B$$.

QED

Like I said, very straightforward algebra, but a little surprising as a result. I wasn’t thinking about the lower tail, so I expected there to be cases where a lower mean lower standard deviation was dominated, but it turns out not.

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