Goldman Sachs Interview Question: What does a distribution with... | Glassdoor

Interview Question

Quantitative Analyst Interview

What does a distribution with a maximal variance look like

  which is only defined between 0 and 1? Give the proof.

Interview Answer

4 Answers


You'd want the values to be at either 0 or 1, since anything in between just brings them closer to the mean (not fully rigorous, but should be obvious enough). let p be the density at 0, so (1-p) is the density at 1. Then mean = (1-p). Variance is: p*(1-p)^2+(1-p)*p^2 = p-2p^2+p^3+p^2-p^3 = p-p^2

We want to maximize variance, to take the first derivative and set it to 0:
1-2p = 0

Hence, half the density is at 0, half at 1.

There may be a more elegant argument.

DW on May 3, 2011

Basically same as DW: ( delta(0) + delta(1) )/2

xeesus on Aug 13, 2012

It is very hard than that to prove the results for given the choice in all probability distributions.

you must find the density f(x) such that:

f is maximising Var(f) = int{0->1} x^2f(x) dx - (int{0->1} xf(x) dx)^2
under the constraint int{0->1} f(x) dx = 1

writing optimality conditions you get that f must be infinite in 0 and 1 and you can use the symmetry argument to get the DW result.

Behe on Nov 10, 2012

Writing optimality does not quite work, since there is an extra constraint of f(x) being positive.
To make DW's answer more rigorous just not that if m is the mean of the distribution, then E((x-m)^2) <= (1-m)^2 (1-m) + m^2 m, since the maximal value of (x-m)^2 above m is (1-m)^2 and below m is m^2. The function (1-m)^3+ m^3 has a maximal value of 1/4, when m = 1/2.

Nerses on Jun 10, 2016

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