The reason is the following. We use the notation:
$$\theta x^i:=\theta_0+\theta_1 x^i_1+\dots+\theta_p x^i_p.$$
Then
$$\log h_\theta(x^i)=\log\frac{1}{1+e^{-\theta x^i} }=-\log ( 1+e^{-\theta x^i} ),$$ $$\log(1- h_\theta(x^i))=\log(1-\frac{1}{1+e^{-\theta x^i} })=\log (e^{-\theta x^i} )-\log ( 1+e^{-\theta x^i} )=-\theta x^i-\log ( 1+e^{-\theta x^i} ),$$ [ this used: $ 1 = \frac{(1+e^{-\theta x^i})}{(1+e^{-\theta x^i})},$ the 1's in numerator cancel, then we used: $\log(x/y) = \log(x) - \log(y)$]
Since our original cost function is the form of:
$$J(\theta)=-\frac{1}{m}\sum_{i=1}^{m}y^{i}\log(h_\theta(x^{i}))+(1-y^{i})\log(1-h_\theta(x^{i}))$$
Plugging in the two simplified expressions above, we obtain
$$J(\theta)=-\frac{1}{m}\sum_{i=1}^m \left[-y^i(\log ( 1+e^{-\theta x^i})) + (1-y^i)(-\theta x^i-\log ( 1+e^{-\theta x^i} ))\right]$$, which can be simplified to:
$$J(\theta)=-\frac{1}{m}\sum_{i=1}^m \left[y_i\theta x^i-\theta x^i-\log(1+e^{-\theta x^i})\right]=-\frac{1}{m}\sum_{i=1}^m \left[y_i\theta x^i-\log(1+e^{\theta x^i})\right],~~(*)$$
where the second equality follows from
$$-\theta x^i-\log(1+e^{-\theta x^i})=
-\left[ \log e^{\theta x^i}+
\log(1+e^{-\theta x^i} )
\right]=-\log(1+e^{\theta x^i}). $$ [ we used $ \log(x) + \log(y) = log(x y) $ ]
All you need now is to compute the partial derivatives of $(*)$ w.r.t. $\theta_j$. As
$$\frac{\partial}{\partial \theta_j}y_i\theta x^i=y_ix^i_j, $$
$$\frac{\partial}{\partial \theta_j}\log(1+e^{\theta x^i})=\frac{x^i_je^{\theta x^i}}{1+e^{\theta x^i}}=x^i_jh_\theta(x^i),$$
the thesis follows.
Here I will prove the below loss function is a convex function.
\begin{equation}
L(\theta, \theta_0) = \sum_{i=1}^N \left( - y^i \log(\sigma(\theta^T x^i + \theta_0))
- (1-y^i) \log(1-\sigma(\theta^T x^i + \theta_0))
\right)
\end{equation}
Then will show that the loss function below that the questioner proposed is NOT a convex function.
\begin{equation}
L(\theta, \theta_0) = \sum_{i=1}^N \left( y^i (1-\sigma(\theta^T x^i + \theta_0))^2
+ (1-y^i) \sigma(\theta^T x^i + \theta_0)^2
\right)
\end{equation}
To prove that solving a logistic regression using the first loss function is solving a convex optimization problem, we need two facts (to prove).
$
\newcommand{\reals}{{\mathbf{R}}}
\newcommand{\preals}{{\reals_+}}
\newcommand{\ppreals}{{\reals_{++}}}
$
Suppose that $\sigma: \reals \to \ppreals$ is the sigmoid function defined by
\begin{equation}
\sigma(z) = 1/(1+\exp(-z))
\end{equation}
The functions $f_1:\reals\to\reals$ and $f_2:\reals\to\reals$ defined by $f_1(z) = -\log(\sigma(z))$ and $f_2(z) = -\log(1-\sigma(z))$ respectively are convex functions.
A (twice-differentiable) convex function of an affine function is a convex function.
Proof) First, we show that $f_1$ and $f_2$ are convex functions. Since
\begin{eqnarray}
f_1(z) = -\log(1/(1+\exp(-z))) = \log(1+\exp(-z)),
\end{eqnarray}
\begin{eqnarray}
\frac{d}{dz} f_1(z) = -\exp(-z)/(1+\exp(-z)) = -1 + 1/(1+exp(-z)) = -1 + \sigma(z),
\end{eqnarray}
the derivative of $f_1$ is a monotonically increasing function and $f_1$ function is a (strictly) convex function (Wiki page for convex function).
Likewise, since
\begin{eqnarray}
f_2(z) = -\log(\exp(-z)/(1+\exp(-z))) = \log(1+\exp(-z)) +z = f_1(z) + z
\end{eqnarray}
\begin{eqnarray}
\frac{d}{dz} f_2(z) = \frac{d}{dz} f_1(z) + 1.
\end{eqnarray}
Since the derivative of $f_1$ is a monotonically increasing function, that of $f_2$ is also a monotonically increasing function, hence $f_2$ is a (strictly) convex function, hence the proof.
Now we prove the second claim. Let $f:\reals^m\to\reals$ is a twice-differential convex function, $A\in\reals^{m\times n}$, and $b\in\reals^m$. And let $g:\reals^n\to\reals$ such that $g(y) = f(Ay + b)$. Then the gradient of $g$ with respect to $y$ is
\begin{equation}
\nabla_y g(y) = A^T \nabla_x f(Ay+b) \in \reals^n,
\end{equation}
and the Hessian of $g$ with respect to $y$ is
\begin{equation}
\nabla_y^2 g(y) = A^T \nabla_x^2 f(Ay+b) A \in \reals^{n \times n}.
\end{equation}
Since $f$ is a convex function, $\nabla_x^2 f(x) \succeq 0$, i.e., it is a positive semidefinite matrix for all $x\in\reals^m$. Then for any $z\in\reals^n$,
\begin{equation}
z^T \nabla_y^2 g(y) z = z^T A^T \nabla_x^2 f(Ay+b) A z
= (Az)^T \nabla_x^2 f(Ay+b) (A z) \geq 0,
\end{equation}
hence $\nabla_y^2 g(y)$ is also a positive semidefinite matrix for all $y\in\reals^n$ (Wiki page for convex function).
Now the object function to be minimized for logistic regression is
\begin{equation}
\begin{array}{ll}
\mbox{minimize} &
L(\theta) = \sum_{i=1}^N \left( - y^i \log(\sigma(\theta^T x^i + \theta_0))
- (1-y^i) \log(1-\sigma(\theta^T x^i + \theta_0))
\right)
\end{array}
\end{equation}
where $(x^i, y^i)$ for $i=1,\ldots, N$ are $N$ training data. Now this is the sum of convex functions of linear (hence, affine) functions in $(\theta, \theta_0)$. Since the sum of convex functions is a convex function, this problem is a convex optimization.
Note that if it maximized the loss function, it would NOT be a convex optimization function. So the direction is critical!
Note also that, whether the algorithm we use is stochastic gradient descent, just gradient descent, or any other optimization algorithm, it solves the convex optimization problem, and that even if we use nonconvex nonlinear kernels for feature transformation, it is still a convex optimization problem since the loss function is still a convex function in $(\theta, \theta_0)$.
Now the new loss function proposed by the questioner is
\begin{equation}
L(\theta, \theta_0) = \sum_{i=1}^N \left( y^i (1-\sigma(\theta^T x^i + \theta_0))^2
+ (1-y^i) \sigma(\theta^T x^i + \theta_0)^2
\right)
\end{equation}
First we show that $f(z) = \sigma(z)^2$ is not a convex function in $z$. If we differentiate this function, we have
\begin{equation}
f'(z) = \frac{d}{dz} \sigma(z)^2 = 2 \sigma(z) \frac{d}{dz} \sigma(z)
= 2 \exp(-z) / (1+\exp(-z))^3.
\end{equation}
Since $f'(0)=1$ and $\lim_{z\to\infty} f'(z) = 0$ (and f'(z) is differentiable), the mean value theorem implies that there exists $z_0\geq0$ such that $f'(z_0) < 0$. Therefore $f(z)$ is NOT a convex function.
Now if we let $N=1$, $x^1 = 1$, $y^1 = 0$, $\theta_0=0$, and $\theta\in\reals$, $L(\theta, 0) = \sigma(\theta)^2$, hence $L(\theta,0)$ is not a convex function, hence the proof!
However, solving the non-convex optimization problem using gradient descent is not necessarily bad idea. (Almost) all deep learning problem is solved by stochastic gradient descent because it's the only way to solve it (other than evolutionary algorithms).
I hope this is a self-contained (strict) proof for the argument. Please leave feedback if anything is unclear or I made mistakes.
Thank you. - Sunghee
Best Answer
Just to remove a little of indices, let me consider the case with $m=1$ and $x,y$ are scalars and not vectors. In this case $J$ becomes:
$$J(\theta,x,y) = -y \log(h_\theta(x)) - (1-y) \log(1-h_\theta(x))$$
$$\frac{\partial h_\theta}{\partial \theta}(\theta,x) = \frac{\partial}{\partial \theta}((1+e^{-\theta x})^{-1}) = x e^{-\theta x} h_\theta^2(x)$$
$$\frac{\partial J}{\partial \theta}(\theta,x,y) = -y\frac{x e^{-\theta x} h_\theta^2(x)}{h_\theta(x)} + (1-y)\frac{x e^{-\theta x} h_\theta^2(x)}{1-h_\theta(x)} = x e^{-\theta x} h_\theta^2(x)(\frac{1-y}{1-h_{\theta}(x)}-\frac{y}{h_\theta(x)})=$$
$$= \frac{x e^{-\theta x} h_\theta(x)}{1-h_\theta (x)}((1-y)h_\theta (x) - y(1-h_\theta (x))) = \frac{x e^{-\theta x} h_\theta(x)}{1-h_\theta (x)} (h_\theta(x) - y)$$
Noticing that $e^{-\theta x} h_\theta(x) = 1-h_\theta(x)$, $\frac{\partial J}{\partial \theta}(\theta,x,y) = x(h_\theta(x) -y)$.