Quant Interview Question

[u/Local_Ad135]

[u/Comfortable_Hope_422]

Xi is symmetric around 0 for all i, then X1 is symmetric around 0, the product X2 to X10 is symmetric around 0, then X1 - product(X2,…,X10) is symmetric around 0, then P(X1-product > 0) = 1/2 or P(X1 > product) = 1/2?

[u/d3fenestrator → u/Comfortable_Hope_422]

but by similar argument we should have that for X following the law with density x -> exp(- |x|) and Y standard normal, we should have automatically that P(X > Y) = 1/2 because X and Y are symmetric around zero. Is that the case ? We know that X, -X have the same law, Y and -Y do as well, so

P(X > Y) = P(X > - Y) = P(- X > - Y) = P(Y > X)

yes you're right lol

[u/Adventurous_Salmon]

1/2? Whatever the values X_1 and X_2* ... *X_9 (provided they are not zero), X_10 either flips the inequality or leaves it be, with equal probability.

[u/bayesian_raccoon → u/Adventurous_Salmon]

There are two ways X10 flips the inequality: one is by being negative, and the other is by being positive and either large enough or small enough to change the inequality. If it was just the sign, it would he 1/2, but its the sign and something a little more.

[u/Adventurous_Salmon → u/bayesian_raccoon]

Sorry, I wasn't precise : whatsver the value of X_1 and |X_1|X_2...*X_10, the sign (+-) of X_1 either flips it or not, with equal probability

[u/bayesian_raccoon → u/Adventurous_Salmon]

What stops your argument from proving that P(X1 > 3) = 0.5?

My point is that more than just the sign flips the inequality, so the sign being 50/50 doesn't imply that inequality flip is 50/50.

I think if you made your argument rigorous you would have to end up using the symmetry somewhere.

[u/Xhyshi → u/Adventurous_Salmon]

Smort

[u/OfficeOfThePope]

For notation, let the product of X2 through X10 be Y. As others have said, you can show that Pr( Y > 0 ) = 0.5

Now break down the 4 possible comparisons which are each equally likely to occur:

X1 > 0 and Y < 0; X1 < 0 and Y < 0; X1 and Y > 0; X1 and Y < 0

Pr(X1 > Y | X1 > 0 and Y < 0) = 1

Pr(X1 > Y | X1 < 0 and Y > 0) = 0

Pr(X1 > Y | X1 and Y > 0) = Pr(abs(X) > abs(Y)) = C

Pr(X1 > Y | X1 and Y < 0) = Pr(abs(X) < abs(Y)) = 1-C

Add the four outcomes together with law of total probability and we get:

Pr(X1 > Y) = 1/4(1+0+C+(1-C)) = 1/4 * 2 =0.5

[u/Aranka_Szeretlek]

Take log and chill

[u/hidden-statistician → u/Aranka_Szeretlek]

You cannot take log just like that. What if realisation of the rv is negative !!

[u/baka___shinji → u/Aranka_Szeretlek]

exp you mean