Given log X ~ N(0,1). Compute the expectation of X.

exp(mu + (sigma^2)/2) = exp(0+1/2) = exp(1/2)

Expanding on the correct answers above: E[X] = E[exp(logX)], and logX is normally distributed. So: E[X} is the moment-generating-function (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).

Complete the square in the integral

When a log of a random variable X is normal, then the X itself is [log-normal](https://en.wikipedia.org/wiki/Log-normal_distribution), right? So if we assume Z = log(X), with mean \miu and standard deviation \sigma (Z ~ N(\miu, \sigma), X can be defined as exp(\miu + \sigma Z). Based on the properties of log-normal distribution, expected value of X would be E[X] = exp(\miu + 0.5*\sigma^2) = exp(0 + 1/2) = exp(1/2).

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(y-1/2)*d(y-1/2) ) = Exp(1/2) So, mojo's ans is correct.

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

P(logX P(X

Sorry misread the problem. ignore.

X has a log-normal distribution, so yes the mean is exp(mu+sigma^2/2)=exp(1/2)

Do they care if you explain the theory or not? I just looked at it, it's standard normal, therefore x=50%

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

This is a basic probability question.

Exp[1/4]