# Expected Value of Continuous Random Variable over Finite Limits

We know that for a Continuous Random Variable under a Normal Distribution $$X \sim \mathcal{N}(\mu,\,\sigma^{2})$$, the Expected Value and the Probability Density Function are as follows:

• $$E(X) = \int_{-\infty}^{\infty} x f(x) dx$$ where $$f(x)$$ is probability density function. (1).
• PDF:

Suppose I have a small dataset (just for simplicity sake):

$$S = [1.1, 2.2, 3.3, 4.4, 5.5, 6.6] ~ where ~ \mu = 3.85 , \sigma=1.87$$

What is the expected value of the above continuous random variable $$X$$ over a finite limits from 1.1 to 6.6?

As far as I know, it is simply the mean $$\mu = 3.85$$ of $$S$$. If not, Do I need to use (1) to integrate over 1.1 to 6.6? if yes, How please?

• What exactly do you wish to compute? $\mathbb E(X\mid 1.1\le X \le 6.6)$ where $X\sim\mathcal N(3.85, 1.87^2)$; or $\mathbb E(X)$ where $X$ is discretely distributed according to your sample? – Maximilian Janisch Oct 17 '19 at 12:23
• @MaximilianJanisch The first for sure please – Mike Oct 17 '19 at 12:27

First of all, assuming that $$\mu$$ and $$\sigma^2$$ are already specified, if you have a random variable $$X \sim \mathcal{N}(\mu, \sigma^2)$$ and you wish to truncate $$X$$ so that its support is limited to the interval $$(1.1, 6.6)$$, if we let $$Y = X \mid (1.1 < X < 6.6)$$ (read as "$$X$$ given $$X$$ is greater than $$1.1$$ and less than $$6.6$$"), then you would have to compute the PDF of $$Y$$, which would be $$f_{Y}(y) = \dfrac{f_{X}(y)}{\mathbb{P}(1.1 < X < 6.6)} = \dfrac{f_{X}(y)}{\Phi\left( \dfrac{6.6-\mu}{\sigma}\right) - \Phi\left( \dfrac{1.1-\mu}{\sigma}\right)}\text{ for } 1.1 < y < 6.6$$ where $$f_{X}(y) = \dfrac{1}{\sigma \sqrt{2\pi}}e^{-(y-\mu)^2/(2\sigma^2)}$$ and $$\Phi(x) = \int_{-\infty}^{x}\dfrac{1}{\sqrt{2\pi}}e^{-t^2/2}\text{ d}t$$ is the $$\mathcal{N}(0, 1)$$ (standard normal) cumulative distribution function.
Thus, $$\mathbb{E}[Y] = \int_{1.1}^{6.6}yf_{Y}(y)\text{ d}y = \int_{1.1}^{6.6}y \cdot \dfrac{\dfrac{1}{\sigma \sqrt{2\pi}}e^{-(y-\mu)^2/(2\sigma^2)}}{\Phi\left( \dfrac{6.6-\mu}{\sigma}\right) - \Phi\left( \dfrac{1.1-\mu}{\sigma}\right)}\text{ d}y\text{.}$$ This will likely have to be computed using numerical approximation.
Second of all, you have a data set and you seem to suggest estimating $$\mu$$ and $$\sigma^2$$ using this data set. There is no one "correct" way to estimate $$\mu$$ and $$\sigma^2$$ with a provided data set. Two estimation procedures include the method of moments (MOM) and the method of maximum likelihood (MLE). Depending on the method you choose, your values for $$\mu$$ and $$\sigma$$ may vary. However, in the case of both of these methods, for the non-truncated normal distribution, the estimators for $$\mu$$ and $$\sigma^2$$ are identical (see MOM, MLE).