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.
Answer

Interview Answer

4 Answers

2

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
p=1/2

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

There may be a more elegant argument.

DW on May 3, 2011
0

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

xeesus on Aug 13, 2012
0

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
0

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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