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

Question

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:

Link:

1. https://www.cs.utexas.edu/~ecprice/courses/randomized/notes/lec2.pdf
2. https://www.probabilitycourse.com/chapter6/6_2_3_chernoff_bounds.php
3. http://crypto.stanford.edu/~blynn/pr/chernoff.html

Thanks.

Answer 1

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 $\lambda>0$.)

This is just a consequence of the fact that (1) multiplication by a positive number is monotone and (2) exponentiation is monotone.