Let $X$ denote a normal random variable and $f$ some differentiable function.

In single-variable calculus, I can prove that :


by integrating by parts.

Indeed, $\mathbb{E}[Xf(X)]=\int_{-\infty}^{+\infty}xf(x)w(x)dx$ with $w(x)=\frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac12 \frac{x^2}{\sigma^2}}$

so it suffices to notice that $\frac{dw(x)}{dx}=-\frac{x}{\sigma^2}w(x)$ to complete the integration by parts and obtain :


Now, some exercise says it can be generalized to vectors :

$\mathbb{E}[X_if(X_1,...,X_n)]=\Sigma_{j=1}^{n}\mathbb{E}[X_iX_j]\mathbb{E}[\frac{\partial f}{\partial X_j}(X_1,...X_n)]$ $\space \space \space$(*)

As far as I know, the probability density function for multinormal random variables is usually defined as $w(\vec X)=Ne^{-\vec x^TC\vec x}$ with $N$ the normalization factor and C the covariance matrix whose inverse yields the $\sigma_{ij}$.

It's been far too long since the last time I studied multivariable calculus (which was barely formally taught to me actually, $2$-$3$ maths exercises max for my whole life as a graduate physics student, rest was just knowing by heart $r^2sin(\theta)$ without ever needing the formal definition of a Jacobian), so now I am having trouble...

What I have tried :

I first notice that in the RHS of (*), the vector differential $\frac{\partial}{\partial \vec Y}$ is involved on $f(\vec X)$, with $\vec Y=(c_{i1}X_1,c_{i2}X_2,...,x_{in}X_n)$ (should be a column vector actually)

Naively generalizing integration by parts, I get the following :

$\int_{-\infty}^{+\infty}\frac{\partial f(\vec X)}{\partial \vec Y}w(\vec X)d\vec X=\color{blue}{\int_{-\infty}^{+\infty}\frac{\partial (f(\vec X)w(\vec X))}{\partial \vec Y}d\vec X} - \int_{-\infty}^{+\infty}f(\vec X)\frac{\partial w(\vec X)}{\partial \vec Y}d\vec X$

If I remember correctly, $\frac{d\vec X}{d \vec Y}$ is the Jacobian matrix, and since $\vec Y$ is just $\vec X$ times a diagonal matrix with diagonal entries $(c_{ij})_{j=1,...,n}$, I have that $\frac{d\vec X}{d \vec Y}=(c_{ij}^{-1})_{j=1,...,n}=(\sigma_{ij})_{j=1,...,n}$

So the blue part is just $\sum_{i=1}^{n} \sigma_{ij}[f(X_i)w(X_i)]_{-\infty}^{+\infty}=0$

Therefore, I am left with :

$\int_{-\infty}^{+\infty}\frac{\partial f(\vec X)}{\partial \vec Y}w(\vec X)d\vec X=- \int_{-\infty}^{+\infty}f(\vec X)\frac{\partial w(\vec X)}{\partial \vec Y}d\vec X$

and I have to prove that :

$- \int_{-\infty}^{+\infty}f(\vec X)\frac{\partial w(\vec X)}{\partial \vec Y}d\vec X = \mathbb{E}[X_if(X_1,...,X_n)] = \int_{-\infty}^{+\infty}X_if(\vec X)w(\vec X)d \vec X$

which could be done if :

$-\frac{\partial w(\vec X)}{\partial \vec Y} = X_iw(\vec X)$

Since $w(\vec X)$ is supposed to be a scalar, I get that :

$-\frac{\partial w(\vec X)}{\partial \vec Y} =-\sum_{j=1}^n \sigma_{ij} \frac{dw(\vec X)}{dX_j}$

Using $\frac{dw(\vec X)}{dX_j}=-w(\vec X)\sum_{i=1}^n c_{ij}X_i$

I obtain :

$-\frac{\partial w(\vec X)}{\partial \vec Y} =w(\vec X)\sum_{k=1}^{n}\sum_{j=1}^n \frac{\sigma_{ij}}{\sigma_{kj}}X_k=w(\vec X)\sum_{k=1}^n X_k \sum_{j=1}^{n}\frac{\sigma_{ij}}{\sigma_{kj}}$

$=w(\vec X)X_i+\color{green}{w(\vec X) \sum_{k=1, k\neq i}^n X_k \sum_{j=1}^{n}\frac{\sigma_{ij}}{\sigma_{kj}}}$

which isn't what I am looking for since the green part doesn't seem to be $0$... and doesn't seem to vanish either when put into the integral as we don't know anything about $f$.

Can someone pinpoint where I am not using multivariable calculus correctly ?

  • $\begingroup$ Would you please tell me how you used integration by parts for $xf(x)w(x)$? $\endgroup$ – zoli Mar 6 '17 at 23:59
  • $\begingroup$ @zoli Yes of course, $\mathbb{E}[Xf(X)]=\int_{-\infty}^{+\infty}xf(x)w(x)dx=-\sigma^2\int_{-\infty}^{+\infty}f(x)\frac{dw(x)}{dx}dx=0+\sigma^2\int_{-\infty}^{+\infty}w(x)\frac{df(x)}{dx}dx=\mathbb{E}[X^2]\mathbb{E}[f'(X)]$ $\endgroup$ – Evariste Mar 7 '17 at 0:10
  • $\begingroup$ The zero comes from the fact that $f(x)w(x)$ vanishes at infinity. (had to split into two comments, because I can't figure out how to skip lines in them and when I don't, it tends to blend in with LateX... $\endgroup$ – Evariste Mar 7 '17 at 0:13

Sorry I am super tired and not completely following your notation so I can't exactly tell you just now where what you're doing is going wrong, but if I offer some pointers as to how this result could be derived which I hope will help.

Note that if $\varphi(\textbf{x})$ is the density of a multivariate normal with covariance matrix $\Sigma$ then from the logarithmic derivative we have $$ \nabla \varphi(\textbf{x}) = - \left(\Sigma^{-1}\textbf{x}\right)\varphi(\textbf{x}), $$ now let $\Lambda = \Sigma^{-1}$, so that an individual component is just given by $$ \frac{\partial \varphi}{\partial x_i} = - \left(\sum_j \Lambda_{ij} x_j \right) \varphi(\textbf{x}), $$ now taking arbitrary $f$ and assuming appropriate boundary conditions we get $$ \begin{align} \int_{\mathbb{R}^n} \frac{\partial f(\textbf{x}) }{\partial x_i} \varphi(\textbf{x})dx &=\int_{\mathbb{R}^{n-1}} \left[\int\frac{\partial f}{\partial x_i} \varphi(\textbf{x}) dx_i \right]d\textbf{x}_{-i} \\ &=\int_{\mathbb{R}^{n-1}}\left[ - \int f(\textbf{x})\frac{\partial \varphi(\textbf{x})}{\partial x_i}\right] d\textbf{x}_{-i} \\ &= \int_{\mathbb{R}^{n-1}}\left[ \int f(\textbf{x})\sum_{j} \Lambda_{ij}x_j\varphi(\textbf{x})dx_i\right] d\textbf{x}_{-i} \\ &= \int_{\mathbb{R}^n} f(\textbf{x}) \left(\sum_j \Lambda_{ij} x_j \right) \varphi(\textbf{x}) d\textbf{x} \end{align} $$ So in terms of expectations and the inverse covariance (precision matrix) we get, $$ \mathbb{E}\left[ \frac{\partial f}{\partial x_i} (\textbf{X}) \right]=\sum_{j}\Lambda_{ij} \mathbb{E}\left[ X_j \cdot f(\textbf{X})\right]. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.