Understanding the pdf of a truncated normal distribution 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?
 A: 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.
