# Why $Pr[X-\mu \geq t]= Pr[e^{\lambda(X-\mu)} \geq e^{\lambda t}]$ for all $\lambda> 0$

I hope everyone is having a nice day. I don't know why this inequality holds.

$$Pr[X-\mu \geq t]= Pr[e^{\lambda(X-\mu)} \geq e^{\lambda t}]$$

For $$\lambda >0$$. I guess it has something to do because the transformation of $$e^x$$ doesn't affect the inequality, but my question is, if that happens, how I know is the same probability?

I am trying to solve this question because I am learning about differential privacy that uses the Chernoff Bound, which uses this equality. These are the links where they use this equality:

Thanks.

The probability is equal because the transformations are equivalences. Read your formula with an unknown non-random variable $$x$$ instead of a random variable $$X$$ to convince yourself that this holds:
$$X-\mu\geq t \iff \lambda(X-\mu)\geq \lambda t \iff e^{\lambda(X-\mu)}\geq e^{\lambda t}.$$
(Of course, this requires that $$\lambda>0$$.)