Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose we have the following theorem:

Theorem. Let $f(\cdot)$ be any pdf. Let $\mu$ be any real number and let $\sigma$ be any positive real number. Then $X$ is a random variable with pdf $(1/\sigma)f((x-\mu)/\sigma)$ if and only if there exists a random variable $Z$ with pdf $f(z)$ and $X = \sigma Z + \mu$.

Proof. ($\Rightarrow$): Let $g(x) = (x-\mu)/\sigma$ and let $Z = g(X)$. Then $g^{-1}(z) = \sigma z + \mu$ and $|(d/dz)g^{-1}(z)| = \sigma$. So the pdf of $Z$ is $$f_{Z}(z) = f_{X}(g^{-1}(z)) \left|\frac{d}{dz}g^{-1} z \right| = f(z)$$

so that $\sigma Z+ \mu = X$.

$(\Leftarrow)$: Suppose $g(z) = \sigma z + \mu$. Then $$f_{X}(x) = f_{Z}(g^{-1}(x)) \left|\frac{d}{dx} g^{-1}(x) \right| = f \left(\frac{x- \mu}{\sigma} \right) \frac{1}{\sigma}$$

Since we can prove any statement by contradiction, would is be difficult to proof both parts of this theorem by contradiction?

share|improve this question
I see you've asked four questions on this site in the past day or so but have yet to upvote or accept any of the answers you've received. So you know, it's considered polite on this site to do these things to let the answerers know that they have been helpful. (Click on the little up arrow by an answer to upvote it, and click on the little check mark by an answer to "accept" it as the best answer to a question.) –  Mike Spivey Jan 28 '12 at 19:34
Why would one want to prove a result by contradiction when a direct proof is available and when there is no indication that a proof by contradiction would be any simpler but rather the opposite? –  Did Jan 29 '12 at 12:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.