What do we need to know about height of men and height of women to conclude that John is likely to be higher than Kate?

Question

UPDATE:

Changed IQ to height to make question seem less controversial.

Suppose we read article that says men have higher mean height than females (distributions of both populations are approximately normal. Both distributions are relatively unskewed). We would be tempted to conclude that randomly chosen male John(whose height we don't know) is likely(i.e. with more than 50% chance) to be higher than randomly selected female Kate(whose height we don't know too). But there is a problem - I don't see how we can go to such conclusion mathematically (or alternatively, how could we prove that such conclusion is false). It feels like some important details are missing for such conclusion. What should we additonally know to conclude that John is likely to be higher than Kate?

Distributions are relatively unskewed, so we can rule out possibility that either the minority of really high males or the minority of very short females drastically influence the mean of their respective population.

The simplest case would be if we knew that any male is higher than any female. Then the mean height of males would be higher than the mean height of all females. So it would be one possible answer to my question: if we knew that even the shortest male is higher than the hightest female, then we would be able to conclude that John is surely higher than Kate (so no probability here).

But under less straightforward circumstances (i.e. when it's NOT true that any male is higher than any female) what do we need to reasonably conclude that John is probably higher than Kate? We could try to say something like "John is likely to be higher than Kate because 51% of males are higher than 51% of females". But such approach looks dubious on closer examination because there are different ways that 51% can be formed. We can prioritize taking the highest avaiable males for 51% of males and be taking the shortiest females first when forming 51% of females. In this case we would be able to say that 51% of males are higher than 51% of females EVEN IF the both population are literally identical! Thus conclusion "John is likely to be higher than Kate" would NOT follow from the premise "51% of males are higher than 51% of females".

P.S. I have found video on Khan academy that explains how to calculate probability that random normal variable W is higher than random normal variable M: https://www.khanacademy.org/math/ap-statistics/random-variables-ap/combining-random-variables/v/analyzing-the-difference-in-distributions?modal=1

Answer 1

If $X_1 \sim N(\mu_1, \sigma_1^2)$ and $X_2 \sim N(\mu_2, \sigma_2^2)$ are independent, then their difference follows the distribution $X_1 - X_2 \sim N(\mu_1 - \mu_2, \sigma_1^2 + \sigma_2^2)$. You can then compute $P(X_1 > X_2) = \Phi(\frac{\mu_1 - \mu_2}{\sqrt{\sigma_1^2+\sigma_2^2}})$ where $\Phi$ is the CDF of the standard normal distribution. When $\mu_1 > \mu_2$, the above probability can be anything between $0.5$ and $1$, depending on the values of $\mu_1, \mu_2, \sigma_1^2, \sigma_2^2$.

Edit: Since $X_1 - X_2 \sim N(\mu_1 - \mu_2, \sigma_1^2 + \sigma_2^2)$ we know $Z := \frac{(X_1 - X_2) - (\mu_1 - \mu_2)}{\sqrt{\sigma_1^2 + \sigma_2^2}} \sim N(0, 1)$. So $$P(X_1 > X_2) = P(X_1 - X_2 > 0) = P\left(Z > - \frac{\mu_1 - \mu_2}{\sqrt{\sigma_1^2+\sigma_2^2}}\right) = \Phi\left(\frac{\mu_1 - \mu_2}{\sqrt{\sigma_1^2+\sigma_2^2}}\right).$$

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One crucial bit of information that is missing from your scenario is how the two people are chosen. John and Kate are two people with fixed heightss, so talking about the "chance" that one's height is larger than the other's does not make sense: it either is or it isn't.

However, if you chose John uniformly at random from the population of men, and the distribution of men's height is $N(\mu_1, \sigma_1^2)$, then John's height can be viewed as a normal random variable. Similarly for Kate's height if she is chosen uniformly at random from the population of women. If these two choices are made independently, then you may use the above computation.

In real life, I seriously doubt you have the opportunity to get a uniformly chosen person from the population, so I would not think the above computation would be particularly applicable in your height scenario.

Answer 2

A more general solution...

Let $X =$ the height of a "random" man, and $Y =$ the height of a "random" woman. Assume $X,Y$ are continuous random variables, independent, and $E[X] - E[Y] = \delta > 0$. Under what conditions can we conclude $P(X>Y) > 1/2$?

The case of two Gaussians has been solved exactly by @angryavian, based on the fact that the difference of two Gaussians is a Gaussian. However, I think the result holds more generally.

Theorem: Suppose (i) both $X$ and $Y$ are symmetric about their respectively means, and (ii) both $X$ and $Y$ have continuous support where the PDF strictly $>0$. (This allows Gaussians, triangles, uniform, etc. and also allow $X,Y$ to be of different "types"). Then $P(X > Y) > 1/2$.

Proof: Let $U = X - E[X], V = Y - E[Y]$ and let $p_U, p_V$ be the PDFs. Then symmetry means $p_U(a) = p_U(-a), p_V(b) = p_V(-b)$ for any $a,b \in \mathbb{R}$.

We will first show $P(U>V) = P(U<V)$. Consider the joint distribution $p_{UV}$ and we have:

• $p_{UV}(a, b) = p_U(a) p_V(b) = p_U(-a) p_V(-b) = p_{UV}(-a, -b)$

• $P(U > V) = \int_{b\in \mathbb{R}} \int_{a>b} p_{UV}(a,b) \, da \, db$

• $P(U < V) = \int_{b\in \mathbb{R}} \int_{a<b} p_{UV}(a,b) \, da \, db= \int_{-b\in \mathbb{R}} \int_{-a > -b} p_{UV}(-a, -b) \, d(-a) \, d(-b) = P(U>V)$

Back to the main result:

• $X < Y \iff U + E[X] < V+ E[Y] \iff U-V < E[Y] - E[X] = -\delta < 0$

• Since $X,Y$ have continuous support where PDF $>0$, so do $U, V$ and $U-V$. Therefore, $-\delta < 0 \implies P(U-V \in (-\delta, 0)) =\epsilon > 0$

• Finally $P(X<Y) = P(U - V < -\delta) = P(U-V<0) - \epsilon < 1/2$. QED

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Further note: Condition (ii) is needed. Without it, here is a counter-example:

• $X =$ uniform in the discontinous support $[50, 60] \cup [70, 80]$

• $Y =$ uniform in the range $[63,65]$

• $E[X] = 65 > E[Y] = 64$ and yet $P(X>Y) = P(X \in [70,80]) = 1/2$, i.e. $\not> 1/2$.