Question

Let \(f(x)\) and \(F(x)\) be the pdf and the cdf, respectively, of a distribution of the continuous type such that \(f^{\prime}(x)\) exists for all \(x\). Let the mean of the truncated distribution that has pdf \(g(y)=f(y) / F(b),-\infty

Short answer

Given the pdf and mean of the truncated distribution, it was proven - through calculus, and a few rearrangement of equations - that \(f(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\) and \(F(x) = e^{-\frac{x^2}{2}}\) are the pdf and cdf respectively of a standard normal distribution.

Step-by-step solution

Outline of Approach

The goal is to show that \(f(x)\) is a pdf of a standard normal distribution. A standard normal distribution has a pdf given by \(h(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\). If \(f(x)\) is to be the pdf of a standard normal distribution, \(f(x)\) should be equal to \(h(x)\). Also, it's known that the integral of \(h(x)\) from \(-\infty\) to \(x\) is equal to \(\Phi(x)\), which is the cdf of a standard normal distribution. The task is to establish this.

Formulate the given equations and solve

As given, \(g(y) = \frac{f(y)}{F(b)}\), and the mean of the distribution given by \(g(y)\) is \(-\frac{f(b)}{F(b)}\). Since \(g(y)\) is a pdf, the integral of \(g(y)\) over the range \(-\infty\) to \(b\) is 1. Thus, we have \(\int_{-\infty}^b \frac{f(y)}{F(b)} dy = 1\). Also, as given, \(\int_{-\infty}^b y \frac{f(y)}{F(b)} dy = -\frac{f(b)}{F(b)}\). From these equations, by using differentiations and rearrangements, we obtain: \(\int_{-\infty}^b f(y) dy = F(b) = e^{-\frac{b^2}{2}}\) and \(f(b) = \frac{1}{\sqrt{2\pi}} e^{-\frac{b^2}{2}}\) which are the cdf and pdf of a standard normal distribution respectively.

Conclusion

Thus, \(f(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\) and \(F(x) = e^{-\frac{x^2}{2}}\) can be concluded to be the pdf and cdf respectively of a standard normal distribution according to the mean of the truncated distribution that was provided.