# Statistics: normal distribution, finding truncated mean.

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?

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 You mean the mean of the truncated distribution? If not, what's a truncated mean? – joriki Oct 29 '12 at 15:31 This integral, integrated over its range would not equal 1. So you would also need to apply some kind of scaling factor for the question of "mean" to make any sense. – Simon Hayward Oct 29 '12 at 15:47 When you are truncating, are you "throwing away" the part before $6$ (probably) or the part after $6$? – André Nicolas Oct 29 '12 at 16:13

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.

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 That's not strictly true. – Arkamis Oct 29 '12 at 16:51

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.

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