Interview Question
Quantitative Researcher Interview

CitadelGiven log X ~ N(0,1). Compute the expectation of X.
Interview Answers
12 Answers
exp(mu + (sigma^2)/2) = exp(0+1/2) = exp(1/2)
mojo on
Expanding on the correct answers above: E[X] = E[exp(logX)], and logX is normally distributed. So: E[X} is the momentgeneratingfunction (mgf) of a standard normal distribution, evaluated at 1. The mgf of a normal distribution with mean mu, SD sigma is exp(mu*t + (1/2) * sigma^2 * t^2), now set mu = 0, sigma = 1, t = 1 to get exp(1/2).
CG on
Complete the square in the integral
interviewee on
Suppose the density function of Y is P(y) and the one for X is F(x), it obeys that P(y)*dy = F(x)*dx; then the expectation of X is E(x) = Integral( x*F(x)*dx ) = Integral( Exp(y) * P(y) * dy ); if you plug the gaussian function and standard deviation in, you will find E(x) = Integral( Exp(1/2) * P(y1/2)*d(y1/2) ) = Exp(1/2) So, mojo's ans is correct.
Anonymous on
I m not that sure, as I got E(x) = 4 I substituted log X = y e^y = X ;and e^2y = t and plz do not forget to change the integration limits
Anonymous on
P(logX P(X
Marcus on
Sorry misread the problem. ignore.
Marcus on
X has a lognormal distribution, so yes the mean is exp(mu+sigma^2/2)=exp(1/2)
Anonymous on
Do they care if you explain the theory or not? I just looked at it, it's standard normal, therefore x=50%
LS on
Let Y = log(X), then X = exp(Y) = r(Y), if we call the pdf of X f(X), then E[X] = integral(Xf(X)dX). By variable transformation, f(x) = g(r^1(X))r^1(X))', plug this into E[X] = integral(Xf(X)dX), we get integral( f(y)dy ), which equals to 1
1 on
This is a basic probability question.
Anonymous on
Exp[1/4]
Anonymous on