Average of truncated Normal Distribution

Question

$N$ points are generated independently from Normal Distribution centered at $x=1$, that is $x_i \sim \mathcal{N}(1,1)$. We define: $$ P = \left\{x_i \ |\ \ x_i > 0\right\} $$ As ${N\to\infty}$ what is the average of points in $P$?

It seems like simple question but I'm not managing to formulate it mathematically and solve it.

Answer 1

So you need to find $\lim_{N\to\infty} \dfrac{\sum_{i=1}^NX_i1_{(X_i>0)}}{\sum_{i=1}^N 1_{(X_i>0)}}$.

Divide the numerator and denominator by $N$, then apply SLLN to see that the above limit is $\dfrac{E(X1_{(X>0)})}{P(X>0)}$.

Now, $P(X>0)=1-\Phi(-1)=\Phi(1)$.

Also, $E(X1_{(X>0)})=E((X-1)1_{(X-1>-1})+E(1_{(X-1>-1)})=E(Z1_{Z>-1)}+P(Z>-1)=E(Z1_{(Z>-1)}+\Phi(1)$.

Now $E(Z1_{(Z>-1)})=\int_{-1}^{\infty}\dfrac{1}{\sqrt{2\pi}}z\exp(-\dfrac{z^2}{2})dz$

Now break the integral into two integrals in $(-1,1)$ and $(1,\infty)$. The first integral is $0$ due to the integrand being odd. The other integral evaluates to $\dfrac{1}{\sqrt{2\pi}}e^{-1/2}$ on using the substitution $z^2/2=t$.

Hence, $\int_{-1}^{\infty}\dfrac{1}{\sqrt{2\pi}}z\exp(-\dfrac{z^2}{2})dz=\dfrac{1}{\sqrt{2\pi}}e^{-1/2}$

So your expression is finally $\dfrac{\Phi(1)+\dfrac{1}{\sqrt{2\pi}}e^{-0.5}}{\Phi(1)}$

Answer 2

The average will converge a.s. to the expected value according to the strong law of large numbers.

Let $\phi$ denote the PDF and let $\Phi$ denote the CDF of standard normal distribution.

Then in general:

$$\int_{a}^{b}u\phi\left(u\right)du=\frac{1}{\sqrt{2\pi}}\int_{a}^{b}ue^{-\frac{1}{2}u^{2}}du=\left[-\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}u^{2}}\right]_{u=a}^{u=b}=\phi\left(a\right)-\phi\left(b\right)\tag1$$

Consequently if $U$ has standard normal distribution then:

$$\mathbb{E}\left(U\mid a<U<b\right)=\frac{\int_{a}^{b}u\phi\left(u\right)du}{\int_{a}^{b}\phi\left(u\right)du}=\frac{\phi\left(a\right)-\phi\left(b\right)}{\Phi\left(b\right)-\Phi\left(a\right)}\tag2$$

Setting $U=X-1$ random variable $U$ has standard normal distribution and application of $(2)$ results in:

$$\mathbb{E}\left(X\mid X>0\right)=\mathbb{E}\left(U+1\mid U+1>0\right)=1+\mathbb{E}\left(U\mid-1<U<\infty\right)=1+\frac{\phi\left(-1\right)}{1-\Phi\left(-1\right)}$$