# Why does Chebyshev's inequality yield that the probability of Laplacian noise being bigger than x is bounded like this?

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.

This can be shown without Chebyshev's inequality. The pdf of a Laplace distribution with scale parameter $$b$$ and mean $$0$$ is $$f(x) = \frac{1}{2b}e^{-\frac{|x|}{b}}$$. Using the symmetry of the distribution about $$0$$:
$$\mathbb{P}(|Lap(b)| > c) = 2\mathbb{P}(Lap(b) > c) = 2\int_c^\infty\frac{e^{-\frac{x}{b}}}{2b}dx = e^{-\frac{c}{b}}$$