How to calculate $\frac{\partial^2}{\partial \theta^2} \bigg(\frac{1}{n} \sum_{i=1}^{n} \exp(-Y_i (\theta^T X_i))\bigg)$? I have a learning sample $D_n = f(X_i, Y_i)_{i=1}^n$ where the $X_i$’s are $\mathbb{R}^d$-valued and the $Y_i$’s are $\{-1, 1\}$-valued. 
$$
f(\theta) = \frac{1}{n} \sum_{i=1}^{n} \exp(-Y_i (\theta^T X_i))
$$
Where $\theta \in [-B, +B]$.
I want to calculate the Hessian matrix: $\nabla^2 f(\theta)$.
 A: Lets focus on what is $\frac{d^2}{d\theta_j^2}f(\theta)$.
Firstly you can bring the operator inside of the sum by linearity. The contribution of $\theta_j$ to the power of $e$ in each term is $-Y_i \theta_j (X_i)_j  $, therefore the derivative multiplies by $-Y_i (X_i)_j$.
$(X_i)_j$ means the $jth$ component of $X_i$.
So hence $\frac{d^2}{d\theta_j^2}f(\theta) = \frac{1}{n}∑_{i=1}^n Y_i^2 (X_i)_j^2 \exp(−Y_i(θ^TX_i))$
Similarly, $\frac{d^2}{d\theta_j \theta_k}f(\theta) = \frac{1}{n}∑_{i=1}^n Y_i^2 (X_i)_j 
 (X_i)_k\exp(−Y_i(θ^TX_i))$
A: Let's employ a convention where uppercase latin letters are matrices, lowercase latin are vectors, and lowercase greek letters are scalars. In this convention, the $(X,Y,\theta,f)$ variables in the problem become $(X,y,w,\phi)$ with dimensions
$$\eqalign{
X&\in{\mathbb R}^{d\times n},\quad y\in{\mathbb R}^{n\times 1},\quad w\in{\mathbb R}^{d\times 1},\quad\phi\in{\mathbb R} \cr
}$$
Let's also define some additional variables which will prove convenient.
$$\eqalign{
&b=X^Tw,\quad a=y\odot b=Yb,\quad &e=\tfrac{1}{n}\exp(-a) \cr
&Y={\rm Diag}(y),\quad E={\rm Diag}(e),\quad &XY=Z\in{\mathbb R}^{d\times n} \cr
&y,a,b,e \in {\mathbb R}^{n\times 1} \cr
&Y,E \in {\mathbb R}^{n\times n} \cr
}$$
Write the function and find its gradient in terms of these new variables.
$$\eqalign{
\phi &= 1:e \cr
d\phi &= 1:de = 1:(-e\odot da) = -e:da \cr&= -e:Y^TX^Tdw = -Ze:dw \cr
g=\frac{\partial\phi}{\partial w} &= -Ze \cr
}$$
So that's the gradient vector. To find the hessian matrix, take the differential of $g$.
$$\eqalign{
dg &= -Z\,de = -Z(-e\odot da) = +ZE\,da \cr
   &= ZE(Y^TX^T\,dw) = ZEZ^Tdw \cr
H = \frac{\partial g}{\partial w} &= ZEZ^T \cr
}$$
In the above steps, the $(\exp)$ function is applied elementwise, $(1)$ is used to denote a vector with all elements equal to one, $(\odot)$ denotes the elementwise/Hadamard product, and $(:)$ denotes the trace/Frobenius product, i.e.
$$\eqalign{A:B = {\rm Tr}(A^TB) }$$
Note that since $(Y,E)$ are diagonal matrices, they are also symmetric. And that the symmetry of the latter ensures that the hessian matrix is symmetric, as it must be.
Since $E$ comes from the $\exp()$ function, it is positive definite and has a unique matrix square-root (which is also positive definite). So the hessian can be factored as
$(Z\sqrt{E})\,(Z\sqrt{E})^T$
and this means that $H$ is at least non-negative definite.
