# What does it mean if the median or average of sums is greater than sum of those of addends?

I'm analyzing the distribution of network latency. The median upload time (U) is 0.5s. The median download (D) time is 2s. However, the median total time (for each data point, T = U + D) is 4s.

What conclusions could be drawn knowing that the median of the sum is much greater than the sum of the medians of the addends?

Just out of curiosity for stats, what would it mean if this question replaced median with average?

• FYI, this cannot be true of the mean, because it is linear: $\mathbb E[X + Y] = \mathbb E X + \mathbb E Y$, and the same is true for sample averages. – Dougal Mar 29 '15 at 17:43

Medians are not linear, so there are a variety of circumstances under which something like that (i.e. $\text{median}(X_1)+\text{median}(X_2)<\text{median}(X_1+X_2)$) might happen.
Here's an explicit example: Take $X_1,X_2 \, \stackrel{\text{i.i.d.}}{ \sim} \operatorname{Exp}(1)$. Then $X_1$ and $X_2$ have median $\log(2) \approx 0.693$ so the sum of the medians is less than $1.4$, but $X_1+X_2\sim \operatorname{Gamma}(2,1)$ which has median $\approx 1.678$ (actually $-W_{-1}(-\frac{1}{2 e}) - 1$ according to Wolfram Alpha)