First, we can use the definition of sub-exponential random variables to show that the product XY is sub-exponential with parameter (\sqrt{2}\sigma \tau, \sqrt{2}\sigma \tau) if and only if for all t > 0:

$$P(|XY| > t) \leq 2\exp(-\frac{t^2}{2(\sqrt{2}\sigma \tau)^2})$$

To prove this, we can use the following lemma:

Lemma: If X and Y are independent sub-gaussian random variables with parameters \sigma and \tau, respectively, then for any a, b \in R:

$$P(|aX + bY| > t) \leq 2\exp(-\frac{t^2}{2(\sigma^2|a|^2 + \tau^2|b|^2)})$$

Proof: We can use the fact that sub-gaussian random variables have tails that decay exponentially. Specifically, for any sub-gaussian random variable Z with parameter \alpha, we have:

$$P(|Z| > t) \leq 2\exp(-\frac{t^2}{2\alpha^2})$$

Using this fact, we can write:

$$P(|aX + bY| > t) = P(|X| > \frac{t}{|a|} - \frac{|b|}{|a|}|Y|)$$

By independence, we can condition on one of the variables, say X, to obtain:

$$P(|aX + bY| > t) \leq \int_{-\infty}^\infty P(|X| > \frac{t}{|a|} - \frac{|b|}{|a|}|y|)P_Y(dy)$$

Since X is sub-gaussian with parameter \sigma, we know that P(|X| > s) \leq 2\exp(-\frac{s^2}{2\sigma^2}) for any s > 0. Using this fact, we can bound the integral as follows:

\begin{aligned} \int_{-\infty}^\infty P(|X| > \frac{t}{|a|} - \frac{|b|}{|a|}|y|)P_Y(dy) &\leq \int_{-\infty}^\infty 2\exp(-\frac{(\frac{t}{|a|} - \frac{|b|}{|a|}|y|)^2}{2\sigma^2})P_Y(dy) \ &= 2\int_{-\infty}^\infty \exp(-\frac{t^2}{2\sigma^2|a|^2} + \frac{|b|^2}{2\sigma^2|a|^2} - \frac{y^2}{2\tau^2|b|^2})P_Y(dy) \ &= 2\exp(-\frac{t^2}{2\sigma^2|a|^2} + \frac{|b|^2}{2\sigma^2|a|^2})\int_{-\infty}^\infty \exp(-\frac{y^2}{2\tau^2|b|^2})P_Y(dy) \ &= 2\exp(-\frac{t^2}{2(\sigma^2|a|^2 + \tau^2|b|^2)}) \end{aligned}

This completes the proof of the lemma.

Now, to prove the main result, we can use the lemma with a = \sqrt{2}\sigma and b = \sqrt{2}\tau to obtain:

P(|XY| > t) \