Let $X\sim N(0, I_d)$ be a $d$-dimensional Gaussian random vector. Let $W_1, \ldots, W_k \in \mathbb{R}^d$ be $k$ fixed vectors in general positions. It is clear that $w_i^\top X, \ldots, w_k^\top X$ are jointly Gaussian random variables. Let $$ Y = \max _{i \in [k]} W_i^\top x ,$$ my question is how to compute $\mathbb{E} [ Y\cdot X]$.

My idea is to use a smooth approximation of the max function and the Stein's identity. Let $$ f(y, \alpha ) = \alpha^{-1} \log \left [ \exp(\alpha y_1) + \ldots + \exp( \alpha y_k ) \right ],$$ it is known that $ \left | \max_i y_i - f(y, \alpha) \right | \leq \log k / \alpha$. Then by Stein's identity, we consider \begin{align} \mathbb{E} \left [ f(WX , \alpha) \cdot X \right ] = \mathbb{E} [ \nabla_{X} f(W X , \alpha) ] = \mathbb{E} \left [ \frac{ \sum_{i \in [k]} \exp ( \alpha \cdot W_i ^\top X ) \cdot W_i }{ \sum_{i \in [k]} \exp(\alpha \cdot W_i ^\top X )} \right ]. \end{align} If I naively take $\alpha \rightarrow +\infty$, I would get \begin{align} &\lim_{\alpha\rightarrow +\infty} \mathbb{E} \left [ f(WX , \alpha) \cdot X \right ] = \mathbb{E} \left[ \sum_{j\in[k]} I\left\{ j =\arg\max _{i\in[k]} W_i^\top X \right \} \cdot W_j \right ] \\ &\quad = \sum_{j\in [k]} \mathbb{P}\left (j =\arg\max _{i\in[k]} W_i^\top X \right ) \cdot W_i. \end{align} I wonder whether my naive derivation gets the correct answer and whether such derivation could be made rigorous.

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    $\begingroup$ Is Y is a scalar-valued variable, and if so, why the dot in $E[Y \cdot X]$? As I see it, here is a simpler formulation of the case d=2: Let (x,y) be bivariate normal, and let (t,u) and (v,w) be fixed vectors. What are E[max(tx+uy,vx+wy)x] and E[max(tx+uy,vx+wy)y]? If that's not what you meant, it would help to state the d=2 case in similar vector-free, subscript-free terms. $\endgroup$
    – user44143
    Jul 22, 2017 at 20:40
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    $\begingroup$ Looks correct and completely rigorous (What is so naiive about passing to the limit? The dominated convergence theorem is still valid and integrable majorants are easy to find:-) ). $\endgroup$
    – fedja
    Jul 22, 2017 at 20:54
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    $\begingroup$ @fedja Thanks very much. I suppose the higher order derivatives can be obtained similarly using DCT. $\endgroup$
    – Steve
    Jul 23, 2017 at 11:57
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    $\begingroup$ @MattF. Sorry for the confusion. The dot in $Y\cdot X$ is the scalar product, that is, $Y$ is multiplied to every entry of $X$. Your statement of the $d=2$ case is exactly what I meant. Thanks! $\endgroup$
    – Steve
    Jul 23, 2017 at 12:00

1 Answer 1


In the $d=2$ case the result is explicit and nice.

Let $x$ and $y$ be i.i.d standard normal, and let $(t,u)$ and $(v,w)$ be fixed vectors.

Let $M = \begin{pmatrix} t & u \\ v & w \\ \end{pmatrix}, \text{ and } X = \begin{pmatrix} x \\ y \\ \end{pmatrix} $. We want $E\big[\max(MX)\,X\big]$.

The first coordinate is \begin{align} &E\big[\max(tx+uy,vx+wy)x\big] \\ ={}&E\big[(tx+uy+vx+wy)x+\left|tx+uy-vx-wy\right|x\big]\ /\ 2 \\ ={}&E\big[(tx+uy+vx+wy)x\big]\ /\ 2 \\ ={}&(t+v)\ /\ 2 \end{align} The first equation uses the identity $\max(a,b)=(a+b+|a-b|)/2$. (This is the part specific to $d=2$: are there any analogs with more than two variables being maximized?)

The second equation uses the symmetry of $(x,y)$ and $(-x,-y)$ in the distribution.

The third equation uses the standard normal variance and independence of $X$ and $Y$.

So using similar reasoning for the second coordinate, $$ E\big[\max(MX)\,X\big]= \frac{1}{2} \begin{pmatrix} t+v\\ u+w\\ \end{pmatrix}. $$


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