# Variation of matrix Bernstein inequality

Let $$[X_1, X_2, ..., X_r]$$ be a set of independent $$d_1 \times d_2$$ dimensional random matrices with $$\mathbb{E}(X_i) = 0$$ and $$\|X_i\| \leq B$$ (bounded operator norm).

Introduce the sum of random matrices, $$Z = \sum_{i=1}^r X_i$$

Define matrix variance proxy:

$$\sigma^2 = \max \left( \| \mathbb{E}(ZZ^T)\|, \|\mathbb{E}(Z^TZ) \|\right) .$$

Then Bernstein inequality for rectangular matrices states that:

$$P(\|Z\| \geq t) \leq (d_1+d_2)\exp\left(\frac{-t^2/2}{\sigma^2 + Bt/3}\right).$$

However, I have also seen people use the following version:

$$P(\|Z\| \geq t) \leq 2(d_1+d_2)\exp\left(\frac{-t^2/2}{\sigma^2 + Bt}\right).$$

My question is how these two inequalities are related to each other? Does one imply another or they are equivalent?

Isn't the first inequality just strictly stronger than the second? As $$$$\sigma^2+Bt/3<\sigma^2+Bt\implies \exp\left(\frac{-t^2/2}{\sigma^2+Bt/3}\right)<\exp\left(\frac{-t^2/2}{\sigma^2+Bt}\right),$$$$ and obviously the $$2$$ factor only makes the second inequality worse.