For the truncated normal distribution below:
$$ {f_X(x; σ, a, b)} = \frac{1}{\sigma}\frac{φ(\frac{x-µ}{σ})}{Φ(\frac{b-µ}{σ})-Φ(\frac{a-µ}{σ})} $$ $$ a = 1; b = ∞; σ = 2 $$
I need to calculate the expected value of X for the general µ and for when µ = 2.
I understand that φ and Φ represent the pdf and CDF of a standard normal distribution random variable.
My first question is, when expanding ${f_X(x; σ, a, b)}$, does it equal to below?
$$ {f_X(x; σ, a, b)} = \frac{1}{\sigma}\frac{φ(\frac{x-µ}{σ})}{Φ(\frac{b-µ}{σ})-Φ(\frac{a-µ}{σ})} =\frac{1}{\sigma}\frac{\frac{1}{\sqrt{2π}}e^{-\frac{x^2}{2}}}{\int_{-\infty}^{\frac{b-µ}{σ}}\frac{1}{\sqrt{2π}\sigma}e^{-\frac{(x-µ)^2}{2{\sigma^2}}}dx - \int_{-\infty}^{\frac{a-µ}{σ}}\frac{1}{\sqrt{2π}\sigma}e^{-\frac{(x-µ)^2}{2{\sigma^2}}}dx } =\frac{1}{\sigma}\frac{\frac{1}{\sqrt{2π}}e^{-\frac{x^2}{2}}}{1 - \int_{-\infty}^{\frac{a-µ}{σ}}\frac{1}{\sqrt{2π}\sigma}e^{-\frac{(x-µ)^2}{2{\sigma^2}}}dx } $$
Question 2a: Can this be simplified further?
Question 2b: Should I be expanding it at all?
The formula for finding the expected value of X is:
$$ E[X] = \int_{-\infty}^{\infty}xf(x)dx $$
Question 3: Should I use this to find the expected value of X? ie. $$ \int_{-\infty}^{\infty}x\cdot\frac{1}{\sigma}\frac{\frac{1}{\sqrt{2π}}e^{-\frac{x^2}{2}}}{1 - \int_{-\infty}^{\frac{a-µ}{σ}}\frac{1}{\sqrt{2π}\sigma}e^{-\frac{(x-µ)^2}{2{\sigma^2}}} }dx $$ which seems very complicated.
Question 4: Is there a better way to find the expected value of X?
Apologies if what I am asking is obvious, I am not familiar with using the φ and Φ concepts. Any help is greatly appreciated.
$$\int_A^Bze^{-z^2/2}\frac{dz}{\sqrt{2\pi}}=\frac{1}{\sqrt{2\pi}}(e^{-A^2/2}-e^{-B^2/2}).\ \ (*)$$If $X=\mu+\sigma Z$ with $Z\sim N(0,1)$, $K=\Pr(a<X<b),$ $A=(a-\mu)/\sigma,$ $B=(b-\mu)/\sigma,$ and $X_1\sim \frac{1}{K}N(\mu,\sigma^2)(dx)1_{a<x<b}(x)$ then $$E(X_1)=\frac{1}{K}\int_A^B(\mu+\sigma z)e^{-z^2/2}\frac{dz}{\sqrt{2\pi}}=\mu+\frac{\sigma}{K\sqrt{2\pi}}(e^{-A^2/2}-e^{-B^2/2}).$$