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Suppose I have two random variable $X$ and $Y$ and they are distributed joint normally and $Y$ is truncated above by constant $c$

$$\begin{pmatrix} X \\ Y \end{pmatrix} = TN\left(\underbrace{\begin{pmatrix} \mu_x \\ \mu_y \end{pmatrix}}_{\text{mean}},\underbrace{\begin{pmatrix} \sigma^2_x & \rho\sigma_x\sigma_y \\ \rho\sigma_x\sigma_y & \sigma^2_y \end{pmatrix}}_{\text{variance matrix}},\underbrace{\begin{pmatrix} -\infty \\ -\infty \end{pmatrix}}_{\text{under bound}},\underbrace{\begin{pmatrix} \infty \\ c \end{pmatrix}}_{\text{upper bound}} \right),$$

From this distribution, I wonder if the mean of $X$ is $\mu_x$. From my calculation, the marginal density of $X$ is $$f_X(x) = \frac{\Phi\left(\frac{c-\mu_y-\frac{\rho\sigma_y}{\sigma_x}(x-\mu_x)}{\sqrt{(1-\rho^2)}\sigma_y}\right)}{\Phi((c-\mu_y)/\sigma_y)}\frac{1}{\sigma_x}\phi((x-\mu_x)/\sigma_x), $$ where $\Phi(\cdot)$ is the cdf of standard normal and $\phi$ is the pdf of standard normal.

I am not sure that if $$\mu_x = \int xf_X(x)\,dx $$

Or maybe $$\mathbb{E}[X] = \mu_x - \rho\sigma_x\left(\frac{\phi((c-\mu_y)/\sigma_y)}{\Phi((c-\mu_y)/\sigma_y)}\right) $$

user1292919
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    This is clearly a duplicate of [Multivariate Normal : expectation of X given Y is doubly-truncated](https://stats.stackexchange.com/questions/385423/multivariate-normal-expectation-of-x-given-y-is-doubly-truncated) – Jarle Tufto Jun 06 '21 at 08:51

1 Answers1

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Since $X$ conditional on $Y$ is not truncated, the usual formula for $EX|Y$ holds. Hence, using the law of total expectation, \begin{align} EX&=EEX|Y\\&=E(\mu_x+\rho\frac{\sigma_x}{\sigma_y}(Y-\mu_y))\\&=\mu_x+\rho\frac{\sigma_x}{\sigma_y}(EY-\mu_y). \end{align} Since $Y$ marginally has is univariate truncated normal distribution,
$$ EY=\mu_y - \frac{\phi(\frac{c-\mu_y}{\sigma_y})}{\Phi(\frac{c-\mu_y}{\sigma_y})}\sigma_y. $$ Thus, $$ EX=\mu_x-\rho\sigma_x\frac{\phi(\frac{c-\mu_y}{\sigma_y})}{\Phi(\frac{c-\mu_y}{\sigma_y})}. $$

Jarle Tufto
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