Let $X_1, X_2, \ldots, X_N$ be independent subgaussian random variables, and let $a_1, a_2, \ldots, a_N$ be nonnegative constants.
Consider the random variable $Y = \sum_{i=1}^N a_i X_i$.
I want to Prove that $$ \|Y\|_{\psi_2} \leq C(\max_{i=1}^N \|X_i\|_{\psi_2}) \|a\|_2, $$ for some positive constant $C$.
I suppose I will be using General Hoeffding's inequality
$$ \mathbb{P}(Y \geq t) \leq 2 \exp\left(-\frac{ct^2}{(\max_{i=1}^N \|X_i\|_{\psi_2})^2 \|a\|_2^2}\right), $$
I tried using Markov's inequality on $\mathbb{P}(Y \geq t)$ such that
$$ \mathbb{P}(Y \geq t) \leq \frac{E(e^{Y^2})}{e^{t^2}} $$
and then equate this to the right hand side of General Hoeffding's but I'm not sure how to substitute this in subgaussian norm to get the right form or if I'm not going in the right direction.
We use one of the equivalent conditions for sub-gaussian random variables. For any $\lambda \in \mathbb{R}$,
\begin{align*} E \exp \left(\lambda \sum_{i=1}^N a_i X_i \right) & = \prod_{i=1}^N E \exp (\lambda a_i X_i) \\ & \leq \prod_{i=1}^N \exp (C\lambda^2 a_i^2 \|X_i \|_{\psi_2}^2) \\ & = \exp \left( C \lambda^2 \sum_{i=1}^N a_i^2\|X_i \|_{\psi_2}^2 \right) \end{align*}
The first equality is because of the independence and the inequality is due to one of the equivalent definitions of sub-gaussian random variables.
Finally, note that
$$ \sum_{i=1}^N a_i^2\|X_i \|_{\psi_2}^2 \leq \max_i \|X_i\|_{\psi_2}^2 \sum_{i=1}^N a_i^2 $$