Let $X\in \mathbb{R}$ be a univariate random varible for which it holds that $$ X \sim N(\mu,\sigma^2).$$ where $\mu\in \mathbb{R}$ gives the expected value and $\sigma^2>0$ is the variance. If we are interested in the first two moments of $X$, given that $X$ is smaller some threshold $b$, Wikipedia provides rather simple expressions for this. Define $\beta=\frac{b-\mu}{\sigma}$ and $\phi$ and $\Phi$ denote the density and cumulative distribution function of a standard normal, then it holds that: $$\mathbb{E}[X|X>b]=\mu-\sigma\frac{\phi(\beta)}{\Phi(\beta)}$$ $$\mathbb{V}[X|X>b]=\sigma^2\bigg{[}1-\beta\frac{\phi(\beta)}{\Phi(\beta)}-\bigg{(}\frac{\phi(\beta)}{\Phi(\beta)}\bigg{)}^2\bigg{]}.$$

So far so good, but what happens if the random variable $X\in \mathbb{R}^n$ is multivariate normally with

$$ X \sim N(\mu,\Sigma).$$

where $\mu\in \mathbb{R}^n$ is the n-dimensional mean vector and $\Sigma$ is the $n \times n$ covariance matrix. If follows that the density of the multivariate truncated normal with all elements of $X$ being smaller than the threshold $b$ (so $X<b$ is always to be viewed elementwise) is given by $$f_X(X,\mu,\Sigma,b)=\frac{\exp\{-0.5(X-\mu)^T\Sigma^{-1}(X-\mu)\}}{\int_{-\infty}^{b}\exp\{-0.5(X-\mu)^T\Sigma^{-1}(X-\mu)\}}I(X<b) $$ Where $I(X<b)$ is an indicator function being one if $X<b$ and zero else. Now my questions: Is there a simple way to represent the first to moments $X$ as in the univariate case?

I.e. what are $\mathbb{E}[X|X<b]$ and $\mathbb{V}[X|X<b]$? I can probably use the results on the univariate truncated normal for the multivariate expectation (just inserting the elementwise univariate expectations) and the diagonal of $\mathbb{V}[X|X<b]$ can be represented by the univariate variances. But what about the covariances?

  • $\begingroup$ what do you mean ? "what are $\mathbb{E}[X|X<b]$ and $\mathbb{V}[X|X<b]$?" $\endgroup$
    – Masoud
    Commented Apr 3, 2019 at 9:34
  • $\begingroup$ The first expression gives the expectation of the truncated random variables and the second one the corresponding variance-covariance matrix. $\endgroup$
    – Michael L.
    Commented Apr 4, 2019 at 10:42
  • $\begingroup$ can you calculate them by $E(X|X<b)=E(X|A)=\frac{E(X1_A)}{p(A)}=\frac{E(X1_{X<b})}{p(X<b)}$? $\endgroup$
    – Masoud
    Commented Apr 4, 2019 at 10:47
  • $\begingroup$ I am not so sure, I will think about it but it is not clear to me whether it really simplifies the problem. $\endgroup$
    – Michael L.
    Commented Apr 4, 2019 at 10:48
  • $\begingroup$ perhaps it is useful $$E(X|X<b)=E(X|A)=\frac{E(X1_A)}{p(A)}=\frac{E(X1_{X<b})}{p(X<b)}$$ $$=\frac{\int x 1_{x<b} f(x) dx}{\Phi(b)}=\frac{\int_{-\infty}^{b} x f(x) dx}{\Phi(b)}$$ $\endgroup$
    – Masoud
    Commented Apr 4, 2019 at 10:55


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