I am trying to understand this proof of the bounds of Laplacian noise used in a paper on differential privacy.
Given a random variable $Lap\left ( \frac{\Delta f}{\varepsilon } \right )$, apparently the probability that the Laplacian noise is bigger than $x$ is bounded by:
$\mathbb{P}\left [ \left | Lap\left ( \frac{\Delta f}{\epsilon } \right ) \right | > x \right ]= e^{-\frac{x \cdot \epsilon}{\Delta f}}$
The proof for this is that the variant below of Chebyshev's inequality "directly" yields this result:
$\mathbb{P}\left [ \left | X - E\left [ X \right ] \right | \geq k^{2}\right ]\leq \frac{\sigma^{2}}{k^{2}}$
The paper doesn't explain this, since it should be very obvious. It is not obvious to me. This is what I did so far:
$E\left [ X \right ]$ denotes expected value.
If I understand it correctly $E\left [ X \right ] = 0$ for
I also think I understand that $\sigma^{2} = \frac{2\cdot \left ( \Delta f \right )^{2}}{ \epsilon^{2}}$ for $Lap\left ( \frac{\Delta f}{\varepsilon } \right )$
But how do you get from that to this: $e^{-\frac{x \cdot \epsilon}{\Delta f}}$??
My first (and only step):
$\mathbb{P}\left [ \left | Lap\left ( \frac{\Delta f}{\varepsilon } \right ) \right | \geq x\right ]\leq \frac{\frac{2\cdot \left ( \Delta f \right )^{2}}{ \epsilon^{2}}}{x}$
Apparently the rest should be completely obvious, but I am completely stumped.