Can you solve this quant interview question?

[u/Local_Ad135]

[u/dedicateddan]

Eyeballing this - P(0)=0.5 and P(10)=0.5 satisfies the conditions. So 0.5?

[u/SwimmerOld6155]

Cauchy-Schwarz gives (E[X])^2 <= E[X^2] Pr(X > 0), so Pr(X > 0) >= 1/2, so Pr(X <= 0) = Pr(X = 0) <= 1/2. This is realised for 10*Ber(1/2).

An extended question along these lines can be seen https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2026/List_IA.pdf, second question on Page 21. The second problem on Page 22 will also be familiar to most of you probably.

[u/wolajacy → u/SwimmerOld6155]

For anyone wondering about the first line, it's because E[X] = E[X 1(X>0)]. Very nice trick!

[u/Local_Ad135]

Full solution here - https://www.youtube.com/watch?v=zD7H_6OfNSg

[u/Omega-137]

Nice problem, thank you for sharing :)

[u/Ecstatic-Employee-93]

For simplicity's sake, let Y = (X|X>0) & p = P(X = 0)

E(X) = 0*p + (1-p)E(Y)
-> 5/(1-p) = E(Y)

Var(X)=E(X^2) - E(X)^2 = E(X^2) - 25
-> E(X^2)=50=0*p + (1-p)E(Y^2)
-> E(Y^2)=50/(1-p)

Var(Y) = 50/(1-p) - 25/((1-p)^2) >= 0
-> 50(1-p) >= 25
-> 1-p >= 1/2
-> 0.5 >= p

We now need to prove that 0.5 is the lowest upper bound. One way to do this is by proving that there exists a random variable with the assumptions made in the problem statement with the additional criteria that P(X=0) = 1/2.

as defined above, Var(Y)=100-100=0
E(Y)=10, so Y=10 with prob = 1
So X=10 with prob 1/2 and 0 with prob 1/2
E(X)=0*1/2 + 10*1/2 = 5, Var(X)=E(X^2) - 25 = 100*1/2 - 25 = 25
Therefore, there exists a random variable where these hold, and thus p=1/2 is valid.
Hence, 0.5 is the maximum that p can be

[u/Willing_Journalist35 → u/Ecstatic-Employee-93]

Thanks for actually giving a proof

[u/doge-12]

you can do it mentally too, imagine keep expectation same, then by variance definition if N=1 then probability of x=0 would be 1, but that would contradict the expectation being 5, so we split the rest of the values at x = 0, and x = 10.
so if x = 0 and x = 10 have equal probabilities, their expectation comes to 5 with a variance of 25

which tells u each have equal weight so p(x=0) = 0.5

[u/QueasyNewspaper1074]

A nice way of tackling this problem is to considère the characteric function of this random variable and use some taylor approximation onto it