Question

Let \(Y\) have a truncated distribution with pdf \(g(y)=\phi(y) /[\Phi(b)-\Phi(a)]\), for \(a

Short answer

The short answer is that the expectation of the truncated standard normal random variable Y as given, is indeed equal to \([\phi(a)-\phi(b)]/[\Phi(b)-\Phi(a)]\). The evaluation of this expectation used integration by parts in calculus, starting off with the definition of expectation for a continuous random variable.

Step-by-step solution

Understand the pdf of Y

This step is essential to lay a foundation for the problem. It is given that the probability density function of Y, denoted by \(g(y)\), is a truncated standard normal distribution defined as \(g(y)= \frac{\phi(y)}{\Phi(b)-\Phi(a)}\) for \(a<y<b\), and zero elsewhere, where \(\phi(x)\) and \(\Phi(x)\) are the pdf and distribution function of a standard normal distribution respectively.

Begin to Calculate the Expectation

The expectation or expected value of a random variable Y, denoted by \(E(Y)\) is defined as the integral of y times the pdf of Y. That is \(E(Y) = \int_{-\infty}^{\infty} y * g(y)dy\). However, because Y is truncated, we only have to integrate from a to b, so \(E(Y) = \int_{a}^{b} y* \frac{\phi(y)}{\Phi(b)-\Phi(a)} dy\). Now, using integration by parts where \(u=y\), \(dv= \frac{\phi(y)}{\Phi(b)-\Phi(a)}dy\), you obtain \(du=dy\) and \(v= -\frac{\phi(y)}{\Phi(b)-\Phi(a)}\).

Implement Integration by Parts

Integration by parts follows the rule: \(\int u dv = u*v - \int v du\). Implementing each calculated part, \(E(Y) = [-y*\frac{\phi(y)}{\Phi(b)-\Phi(a)}]_a^b + \int_a^b \frac{\phi(y)}{\Phi(b)-\Phi(a)} dy\). The last term is 1 (as it is a pdf and integrates to 1 over its range), hence the entire equation simplifies to \(E(Y) = [\phi(a)-\phi(b)]/[\Phi(b)-\Phi(a)]\), which was what was initially required to prove.