Can we give high-probability exponential bounds on the slope of the linear regression function? Suppose $(X,Y), (X_1,Y_1),(X_2,Y_2),\dots$ is a $\mathbb{P}$-i.i.d. sequence of pairs of real-valued random variables such that the support of $\mathbb{P}_{(X,Y)}$ is contained in the square $[-1,1] \times [-1,1]$.
Assume that there exists a noisy linear relation with slope $m  \in [-1,1]$ between $X$ and $Y$, i.e., $\mathbb{E}[Y \mid X] = mX$.
Our goal is to give a reliable estimate of $m$ having access to a sample of size $T\in \mathbb{N}$, say $S_T := \big((X_1,Y_1), \dots , (X_T,Y_T)\big)$. Our performances will be measured via the square loss, i.e., if we predict $\hat{m}$, we pay the loss
\begin{equation*}
   \ell(\hat{m}) :=\mathbb{E}\big[(Y-\hat{m}X)^2\big]\;,
\end{equation*}
which has its minimum at $\hat{m} = m$ (since $\mathbb{E}[Y \mid X] = mX$).
A viable strategy seems to minimize a proxy of $\ell$, for example its empirical version based on the sample $S_T$ we have access to:
\begin{equation*}
   \hat{\ell}_{S_T}(\hat{m}) := \frac{1}{T} \sum_{t=1}^T(Y_t-\hat{m}X_t)^2\;,
\end{equation*}
whose minimum occurs at
\begin{equation*}
   \hat{m}_{S_T} := \frac{\sum_t^T X_tY_t}{\sum_{t=1}^T X_t^2}\;.
\end{equation*}
I'm interested in high-probability exponential guarantees about the performance of the estimator $\hat{m}_{S_T}$ in the spirit of Hoeffding's inequality. Specifically:

Can we guarantee that there exists constants $c,d\ge0$ such that
\begin{equation*}
   \forall \varepsilon >0, \forall T \in \mathbb{N}, \qquad  \mathbb{P}\big[|\hat{m}_{S_T} - m| \ge \varepsilon \big] \le c \cdot \exp (- d \cdot \varepsilon ^2 \cdot T)
\end{equation*}

If that's true, how do these two constants $c,d$ look like?
If that's not true, which kind of guarantees can we give?
 A: I was thinking about something along the following line. Assume that we know a constant $a \in (0,1)$ for which $\mathbb{E}[X^2] \ge a$. We have
\begin{align*}
   |\hat{m}_{S_T} - m|
&\le
   \Bigg|\frac{\frac{1}{T} \sum_{t=1}^T X_t Y_t}{\frac{1}{T} \sum_{t=1}^T X_t^2} - \frac{\frac{1}{T} \sum_{t=1}^T X_t Y_t}{\mathbb{E}[X^2]} \Bigg| +    \Bigg|\frac{\frac{1}{T} \sum_{t=1}^T X_t Y_t}{\mathbb{E}[X^2]} - m \Bigg|
\\
&\le
   \Bigg|\frac{1}{\frac{1}{T} \sum_{t=1}^T X_t^2} - \frac{1}{\mathbb{E}[X^2]} \Bigg| +    \Bigg|\frac{\frac{1}{T} \sum_{t=1}^T X_t Y_t}{\mathbb{E}[X^2]} - m \Bigg|
\\
&=
\Bigg|\frac{\mathbb{E}[X^2]-\frac{1}{T} \sum_{t=1}^T X_t^2}{\mathbb{E}[X^2]\frac{1}{T} \sum_{t=1}^T X_t^2} \Bigg| +    \Bigg|\frac{1}{T} \sum_{t=1}^T \frac{X_t Y_t}{\mathbb{E}[X^2]} - m \Bigg| \;.
\end{align*}
Now, we know that
\begin{equation*}
   \mathbb{P}\Big[ \frac{1}{T} \sum_{t=1}^T X_t^2 < \mathbb{E}[X^2] - \frac{a}{2}\Big] \le \exp\Big(-\frac{1}{8} a^2 T \Big)
\end{equation*}
Now, if $\frac{1}{T} \sum_{t=1}^T X_t^2 < \mathbb{E}[X^2] - \frac{a}{2}$ we have that
\begin{equation*}
   \Bigg|\frac{\mathbb{E}[X^2]-\frac{1}{T} \sum_{t=1}^T X_t^2}{\mathbb{E}[X^2]\frac{1}{T} \sum_{t=1}^T X_t^2} \Bigg|
\le
   \frac{2}{a^2} \Big| \mathbb{E}[X^2]-\frac{1}{T} \sum_{t=1}^T X_t^2 \Big|
\end{equation*}
and so, for any $\varepsilon \in (0,1)$, using a union bound, we get:
\begin{align*}
   \mathbb{P} &\Big[ |\hat{m}_{S_T} - m| \ge \varepsilon \Big]
\\
&\le
   \exp\Big(-\frac{1}{8} a^2 T \Big) + \mathbb{P}\bigg[ \frac{2}{a^2} \Big| \mathbb{E}[X^2]-\frac{1}{T} \sum_{t=1}^T X_t^2 \Big| \ge \frac{\varepsilon}{2} \bigg] + \mathbb{P}\Bigg[\Bigg|\frac{1}{T} \sum_{t=1}^T \frac{X_t Y_t}{\mathbb{E}[X^2]} - m \Bigg| \ge \frac{\varepsilon}{2}\Bigg]
\\
&\le
   \exp\Big(-\frac{1}{8} a^2 T \Big) +  2\exp\Big(-\frac{a^4}{32} \varepsilon^2 T \Big) + 2\exp\Big(-\frac{a^2}{8} \varepsilon^2 T \Big)
\\
&\le
5 \cdot \exp\Big(-\frac{a^4}{32} \cdot \varepsilon^2 \cdot T \Big) \;.
\end{align*}
However, for this line of thought to work, we need an a priori knowledge of a lower bound $a$ on $\mathbb{E}[X^2]$. Also, the bound depends on $a$ in a dreadful way.
If someone has better ideas, please post other answers.
