Statistics: normal distribution, finding truncated mean.

Question

Say $X$ is a random variable arising from a normal distribution with mean $10$ and variance $4$ $(N(10, 4))$ truncated at $X=6$. How do I find the truncated mean of the distrubution?

Answer 1

There isn't a closed form expression for such a mean, in general. You can estimate it with Monte Carlo methods, by drawing a large sample from the truncated distribution and taking the sample mean.

Answer 2

Approximating one mean-square convergent distribution with another such distribution is very easily done using generalized polynomial chaos.

Say you have a random variable $Y$ that is a truncated (either left, right, or two-way) normal random variable. It is possible to approximate this using a spectral expansion about orthogonal polynomials in a second random variable, $\zeta$, whose distribution we choose. The choice of distribution governs the family of orthogonal polynomials.

In this case, let us choose to model $Y$ with a Gaussian distribution. The polynomial chaos representation is:

$$Y = \sum_{i=0}^\infty y_i \Phi_i(\zeta).$$

Here, $\Phi_i$ represents the $i$th Hermite polynomial, and $\zeta$ is a Gaussian R.V.

Then, the challenge becomes one of computing the coefficients $y_i$. We expect this series to converge (accept that prima facie; the proof is very complex), so we may truncate it at $P$ terms. Then, we apply the Galerkin method $P+1$ times to compute each $y_i$:

$$y_i = \frac{\langle Y \Phi_i(\zeta)\rangle}{\langle \Phi_i^2(\zeta) \rangle}.$$

Since $\Phi$ is orthogonal, the denominator is easy to compute (indeed, we can choose a slightly scaled Gaussian R.V. such that $\langle \Phi_i^2(\zeta) \rangle = 1$).

Then, the numerator may be computed

$$\langle Y \Phi_i(\zeta) \rangle = \int_{-\infty}^\infty Y \Phi_i(\zeta) g(\zeta)\ d\zeta. \tag{1}$$

To compute this integral, we need to cast $Y$ in terms of $\zeta$. Since the image of CDF of both $Y$ and $\zeta$ is the interval $[0,1]$, we cast them to a uniform R.V. using an inverse sample transform. Then, Equation 1 becomes

$$ \int_{-\infty}^\infty Y \Phi_i(\zeta) g(\zeta)\ d\zeta = \int_0^1 h(u)\Phi_i(l(u))\ du$$

where

$$\begin{align*} h(u) &= F^{-1}(Y) \\ l(u) &= G^{-1}(\zeta) \end{align*}$$

and $F^{-1}$ and $G^{-1}$ represent the inverse CDFs of $Y$ and $\zeta$ respectively.

These integrals may be computed numerically, or in some cases they have an elementary closed form.

However, regardless of the basis distribution you choose, the coefficient $y_0$ will represent the mean of $Y$.

This is extremely convenient, because typically first orthogonal polynomial, $\Phi_0(\zeta) = \textrm{const}$. This means that the integral is much more likely to have a nice closed form, either an elementary closed form, or a closed form as a special function.

In fact, it may even be said that the closed form representation of the mean is exactly $\int_0^1 kh(u)\ du$ for the appropriate scaling of your basis R.V.

Here is the result of approximating a Truncated Normal distribution using a Gaussian distribution:

In this case, I specified the mean and variance of the truncated normal distribution, but in principle one can take a slightly different approach for estimating $F^{-1}$ that does not require this information. Alternatively, one can then compute the scaling parameter, as mentioned by @SimonHayward, using this approach.