Expected value of Truncated Normal Distribution [duplicate]

Question

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.

Answer 1

I would provide you a general hint.

What does truncation actually do? For $X\sim f$ having an infinite support, the distribution of $X$ in $(a, b] ,~a,b\in\bar{\mathbb R}$ is $$ \frac{f(x)\cdot\boldsymbol 1_{x\in(a,b]}}{\Pr(a<X\leq b)};$$ the denominator is nothing but $F(b) -F(a). $ When $b=\infty,$ it is $S(a). $

When $X$ is normal, this means we need to evaluate $\mathbb E[X\mid X> a].$

Use $f^\prime(x) =\frac{-(x-\mu) }{\sigma^2}f(x) $ and few simple algebraic manipulations (work with $S$ rather than $F$) to yield $\mu+\sigma^2\frac{f(a) }{S(a) }.$

No need to break the entire pdf in numerator and denominator. Work wisely as it involves nothing but regular calculus procedure and property of normal density.