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:

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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?

  • $\begingroup$ 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? $\endgroup$ – Maximilian Janisch Oct 17 '19 at 12:23
  • $\begingroup$ @MaximilianJanisch The first for sure please $\endgroup$ – Mike Oct 17 '19 at 12:27

You're dealing with multiple issues here - one deals with an expected value calculation given truncation (probability), and another deals with estimation of parameters using data (statistics). Given that you haven't specified a method for estimation of your parameters, my guess is that there's a lot of background that you're missing.

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).


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