Suppose $\boldsymbol{x} = (x_1, \ldots, x_m)^T$ follows a multivariate normal distribution with 2-sided truncation $a_i \leq x_i \leq b_i$. This is a truncated multivariate normal defined by $TN(\mu, \Sigma, a, b)$ where $a = (a_1, \ldots, a_m)^T$ and $b = (b_1, \ldots, b_m)^T$.

The probability density function for $TN(\mu, \Sigma, a, b)$ can be expressed as

$$f(x, \mu, \Sigma, a, b) = \frac{\exp\left\{-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)\right\}}{\int_{a_1}^{b_1} \int_{a_2}^{b_2}\cdots \int_{a_m}^{b_m} \exp\left\{-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)\right\}dx_m \cdots dx_1}$$

for $a \leq x \leq b$ and 0 otherwise.

My question is: suppose $x = (x_1, x_2)$, is it possible to have the following pdf for $TN\left(\mu, \Sigma, a = (a_1, a_2), b = (b_1, 2x_1)\right)$

$$f(x, \mu, \Sigma, a, b) = \frac{\exp\left\{-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)\right\}}{\int_{a_1}^{b_1} \int_{a_2}^{2x_1} \exp\left\{-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)\right\}dx_2dx_1}$$

where the upper truncation for $x_2$ depends on $x_1$? Is this valid?

  • 3
    $\begingroup$ Don't let the notation confuse you: you can truncate any distribution to any measurable set of positive measure. The procedure is the same: all values outside the set get zero probability and the chances of all values inside the set are inflated to make the total truncated probability equal to $1.$ Here's an example of truncation to the complement of a disk: stats.stackexchange.com/questions/157963 and here is an example of arbitrarily complex truncation in one dimension: stats.stackexchange.com/a/491045/919. $\endgroup$
    – whuber
    Commented May 25, 2021 at 18:56
  • $\begingroup$ @whuber, I see. So to double-check, it is valid to have something like $TN(\mu, \Sigma, a = (a_1, a_2), b = (b_1, 2x_1))$ where the truncation for $x_2$ depends on $x_1$? $\endgroup$
    – Adrian
    Commented May 25, 2021 at 18:58
  • 1
    $\begingroup$ I don't understand what that notation means. But unless you are truncating to a vertical rectangle in $(x_1,x_2)$ coordinates, along the boundary of the truncation region $x_2$ must vary with $x_1.$ $\endgroup$
    – whuber
    Commented May 25, 2021 at 19:00
  • $\begingroup$ I'm writing $TN(\mu, \Sigma, a = (a_1, a_2), b = (b_1, 2x_1))$ to denote a truncated bivariate normal distribution where $x_1$ lies between $a_1$ and $b_1$, and $x_2$ lies between $a_2$ and $2x_1$. That is, the truncation region of $x_2$ varies with $x_1$. $\endgroup$
    – Adrian
    Commented May 25, 2021 at 19:02
  • $\begingroup$ Just to understand your comment better @whuber, if I truncated to $a_1 \leq x_1 \leq b_1$, $a_2 \leq x_2 \leq b_2$, this is equivalent to truncating to a rectangle in $(x_1, x_2)$ coordinates. In my case, having $a_1 \leq x_1 \leq b_1, a_2 \leq x_2 \leq 2x_1$ is equivalent to truncating to something other than a rectangle, correct? My question is: does the multivariate truncated normal distribution ALWAYS truncate to a rectangle, i.e., do lower limits $a$ and upper limits $b$ always have to be constants? Thank you. $\endgroup$
    – Adrian
    Commented May 25, 2021 at 19:11

1 Answer 1


Perhaps a more general notation will uncover the basic concepts and help you answer your question. There's little more to the following analysis than using mathematical notation carefully. Because that makes it abstract, I rephrase the key results in English: look for the quoted passages.

When there are $m$ random variables $X=(X_1, \ldots, X_m)$ they have a distribution. This distribution gives the chance that $X$ lies in any Borel measurable set $\mathcal{R},$ written

$$F_X(\mathcal{R}) = \Pr(X\in\mathcal{R}).$$

The distribution is (absolutely) continuous when there is a density function $f_X$ defined on all of $\mathbb{R}^m$ whose integral gives the probability. That is, for all $\mathcal R,$

$$\Pr(X\in\mathcal{R}) = F_X(\mathcal{R})= \iint_\mathcal{R} f_X(x)\mathrm{d}x = \iint_{\mathbb{R}^m} \mathcal{I}_{\mathcal{R}}(x)f_X(x)\,\mathrm{d}x.\tag{*}$$

The latter expression involves the indicator function $\mathcal{I}_{\mathcal{R}}$ (which by definition takes the value $1$ at all points in $\mathbb{R}$ and otherwise is zero). It enables us to express the integral over any region in terms of an integral over the entire space $\mathbb{R}^m.$

Suppose $\mathcal{E} \subset \mathbb{R}^m$ is a measurable set. Then the truncation of $F_X$ to $\mathcal{E}$ is a distribution function that arises when we "throw out all outcomes where $X$ is not in $\mathcal{E}.$" It is obtained in the simplest possible manner: just "limit the probability to the part of $\mathcal{R}$ lying in $\mathcal{E}:$"

$$F_X^{\mathcal{E}}(\mathcal{R})\, \propto\, F_X(\mathcal{E}\cap\mathcal{R}).$$

The implicit multiple $\lambda$ in this proportion has to be such to make the total probability $1,$ leading to the equation

$$1 =F^{\mathcal{E}}_X(\mathbb{R}^m) = \lambda F_X(\mathcal{E}\cap\mathbb{R}^m)= \lambda F_X(\mathcal{E})$$

yielding a unique (and obvious) value for $\lambda$ which we may plug into the foregoing to yield

$$F^{\mathcal{E}}_X(\mathcal{R}) = \frac{F_X(\mathcal{E}\cap\mathcal{R})}{F_X(\mathcal{E})}.$$

This reads as "the chance $X$ is in the part of $\mathcal{R}$ lying in $\mathcal{E}$ relative to the chance $X$ is in $\mathcal{E}.$"

When $F_X$ is absolutely continuous the identity $\mathcal{I}_{\mathcal{E}\cap\mathcal{R}} = \mathcal{I}_{\mathcal{E}}\,\mathcal{I}_{\mathcal{R}}$ (a consequence of the arithmetic facts $1\times1=1$ and $1\times 0 = 0\times 0 = 0$) produces

$$F^{\mathcal{E}}_X(\mathcal{R})\, \propto\, \iint_{\mathcal{E} \cap \mathcal{R}} f_X(x)\,\mathrm{d}x = \iint_{\mathbb{R}^m} \mathcal{I}_{\mathcal{E}}(x)\mathcal{I}_{\mathcal{R}}(x)f_X(x)\,\mathrm{d}x = \iint_{\mathcal{R}} \mathcal{I}_{\mathcal{E}}(x)f_X(x)\,\mathrm{d}x$$

and therefore

$$F^{\mathcal{E}}_X(\mathcal{R}) = \frac{\iint_{\mathcal{R}} \mathcal{I}_{\mathcal{E}}(x)f_X(x)\,\mathrm{d}x}{\iint_{\mathcal{E}} f_X(x)\,\mathrm{d}x} = \iint_{\mathcal{R}}\frac{ \mathcal{I}_{\mathcal{E}}(x)f_X(x)}{\iint_{\mathcal{E}} f_X(x)\,\mathrm{d}x}\,\mathrm{d}x = \iint f_X^{\mathcal{E}}(x)\,\mathrm{d}x.$$

This exhibits the density of the truncated variable as

$$f_X^{\mathcal{E}}(x) = \frac{ \mathcal{I}_{\mathcal{E}}(x)f_X(x)}{\iint_{\mathcal{E}} f_X(x)\,\mathrm{d}x}.$$

It is "the density $f_X,$ zeroed beyond $\mathcal{E},$ as renormalized to integrate to unity."

Here is an application. Let $m=2$ and suppose $\mathcal{E}$ is the region defined by

$$\mathcal{E} = \{(x_1,x_2)\mid a_1\le x_1\le b_1,\, a_2 \le x_2 \le 2x_1\}.$$ It is either empty, a point, a triangle, or (generically) a trapezoid. Applying the preceding analysis shows that for any density $f_{X_1,X_2},$ the truncated density is given by Fubini's Theorem as

$$\begin{aligned} f_{X_1,X_2}^\mathcal{E}(x_1,x_2) &= \frac{ \mathcal{I}_{\mathcal{E}}(x)f_X(x)}{\iint_{\mathcal{E}} f_X(x)\,\mathrm{d}x} \\ &= \frac{ \mathcal{I}_{\mathcal{E}}(x_1,x_2) f_{X_1,X_2}(x_1,x_2)}{\int_{a_1}^{b_1}\int_{a_2}^{2x_1} f_{X_1,X_2}(x_1,x_2)\,\mathrm{d}x_2\mathrm{d}x_1} \\ &= \frac{ f_{X_1,X_2}(x_1,x_2)}{\int_{a_1}^{b_1}\int_{a_2}^{2x_1} f_{X_1,X_2}(x_1,x_2)\,\mathrm{d}x_2\mathrm{d}x_1} \end{aligned}$$

for $(x_1,x_2)\in\mathcal{E}$ (and zero otherwise). When you use the binormal density for $f_X$ you have the answer to the question.


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