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I would like to calculate $E[log{(a-b*x)^2}]$ where $a$ is known scaler, $b$ is a known row vector and $x$ is random column vector following a multivariate normal distribution with a known mean and covariance matrix. What is a quick way to calculate or approximate this expectation numerically [without using simulation or difficult multivariate integration]?

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You want to use the Delta method. The Delta method is based on a Taylor series expansion of $\log(a-bX)^2$ around its mean.

In general, if $X$ is a random variable, then the approximate value of $E(f(X))=f(\mu_X)$. For a univariate random variable $X$, the approximate variance of $f(X)$ is $[f'(\mu_X)]^2 \sigma^2_X$. In the multivariate case, this becomes

$$ f'(\mu)^T \, \Sigma \, f'(\mu)$$

Using this method, the expectation of your function is approximately $\log(a-b*\mu)^2$, where $\mu$ is a column vector and the mean of $X$.

The variance of your transformed variable is $c'\Sigma c$, where $\Sigma$ is the variance of $X$ and element $i$ of $c$ is

$$ \frac{-2b_i}{a-b*\mu}$$

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You say that $a, b, \mu, \Sigma$ are all known and you want a numerical solution. In that case, consider (sticking with your notation where $b$ is a row $p$-vector and $X$, $\mu$ are column $p$-vectors):

$Y = a - bX \sim N_1(a-b\mu, b\Sigma b')$,

that is, $Y$ is univariate normal. At this point you have a couple of choices. One is to recognize that $Y^2$ will have distribution proportional to a non-central $\chi^2_1$ distribution. Do your integration of $\log Y^2$ with respect to that density. I think that's going to be a PITA, myself, because of the form of the noncentral $\chi^2_1$ density.

So I would opt for approach two, which is to brute force the integration of $\log Y^2$ with respect to the relevant Normal density.

Note that while the delta method will give you approximate answers, in this case Jensen's inequality guarantees it will be biased. In that case, you have to worry about mean squared error rather than variance.

I will admit to some curiosity about the context that gives rise to this integral.

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As other answers pointed out, $Y = a-\mathbf b' \mathbf x$ is univariate normal with non-zero mean $\mu_y$ and variance $\sigma^2_y$. Then the random variable

$Z \equiv Y^2/\sigma^2_y$ follows a non-central chi-square distribution with degrees of freedom $k=1$ and non-centrality parameter $\lambda = \mu_y^2/\sigma^2_y$.

So you want to calculate

$$E\left[\ln\big(\sigma^2_yZ\big)\right] = \ln\sigma^2_y+E\left[\ln Z\right]$$

The first term you say you know. As for the second term, in this internet address, you can find a formula for $E\left[\ln Z\right]$ (I have not checked its correctness). The author states that this is a result from his PhD Thesis, and he provides an e-mail address if you want to contact him. I would suggest that you do, in order to "map" correctly his definition of a non-central chi-square to yours (it appears he examines the complex distribution).

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