# What is the expected value of $F_x(a + BX)$ where $X \sim MVN(\mu, \sigma^2I)$, $a, B$ are constants, and $F_x$ is the cdf of $X$?

Suppose $$a \in \mathbb{R}^n$$ is a vector of constants, $$B \in \mathbb{R}^{n \times p}$$ is a matrix of constants, and $$X \sim MVN_p(\mu, \sigma^2I)$$ with cdf $$F_x$$.

I know that $$F_x(X)$$ follows a $$Unif(0, 1)$$ distribution. Now define $$Y = a + BX$$. $$Y$$ is normally distributed with mean $$a + B\mu$$ and covariance $$\sigma^2BB^T$$ because $$Y$$ is a linear transformation of $$X$$.

My question: is $$F_x(Y)$$ still a $$Unif(0, 1)$$ and what is its expected value?

My attempt:

\begin{align*} P(F_x(Y) \leq z) &= P(Y \leq F^{-1}_x(z))\\ &= F_y(F_x^{-1}(z))\\ &\neq z \end{align*}

Therefore, $$F_x(Y) \nsim Unif(0, 1)$$. How can I find its expected value, i.e., $$E[F_y(F_x^{-1}(z))]$$ in closed form?

• Try working in $p=1$ with some concrete values (e.g., $\mu=0$, $\sigma^2=1$, $B=1$, $a=100000$) to try to convince yourself whether the claim holds even in one dimension. Jan 3 at 3:22
• Just tried it and I don't think the claim holds. Since it does not follow a uniform distribution, is it still possible to find its expected value, i.e., $E[F_y(F_x^{-1}(z))]$ in closed form? Jan 3 at 3:33
• You mean you want to find $E[F_x(Y)]$ in closed form? Jan 3 at 7:35
• And how are you inverting a multivariate CDF? Jan 3 at 7:41
• Also, the question makes no sense in general since $Y$ is $n\times 1$ so is not a valid argument in the CDF of $X$ for $n\neq p$. Jan 3 at 7:47

I am not sure what the motivation of the question is, but note that it does not make sense as currently framed for $$n\neq p$$ since $$Y$$ is $$\mathbb{R}^n$$-valued while the domain of the CDF of $$X$$ is $$\mathbb{R}^p$$. Also, in terms of your work, I am not sure how you are inverting a multivariate CDF.
So let's assume $$p=n=1$$. Define $$Z=F_X(a+BX)$$, and assume $$B\neq0$$ (clearly $$Z$$ is not uniform if $$B=0$$). Then its CDF is
$$F_Z(z)=P(F_X(a+BX)\leq z)=F_X\left(\frac{F_X^{-1}(z)-a}{B}\right),$$
$$f_Z(z)=\frac{1}{B}\frac{f_X\left( \frac{F_X^{-1}(z)-a}{B}\right)}{f_X\left(F_X^{-1}(z)\right)},$$
which is clearly not a uniform density (except when $$a=0,B=1$$ as expected). I will let you have fun finding the mean with that density.