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?

  • $\begingroup$ 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. $\endgroup$
    – angryavian
    Jan 3 at 3:22
  • $\begingroup$ 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? $\endgroup$
    – Adrian
    Jan 3 at 3:33
  • $\begingroup$ You mean you want to find $E[F_x(Y)]$ in closed form? $\endgroup$ Jan 3 at 7:35
  • $\begingroup$ And how are you inverting a multivariate CDF? $\endgroup$ Jan 3 at 7:41
  • $\begingroup$ 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 $n\neq p$. $\endgroup$ Jan 3 at 7:47

1 Answer 1


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

$$F_Z(z)=P(F_X(a+BX)\leq z)=F_X\left(\frac{F_X^{-1}(z)-a}{B}\right),$$

implying a density

$$f_Z(z)=\frac{1}{B}\frac{f_X\left( \frac{F_X^{-1}(z)-a}{B}\right)}{f_X\left(F_X^{-1}(z)\right)},$$

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.


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