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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