Let X be a real-valued random variable with density f . Further, assume that there exists µ ∈ R such that f (µ+ x) = f (µ− x), for all x ∈ R.

Show that the random variables (µ − X) and (X − µ) have the same distribution, i.e., have the same cdf.

I know that two random variables have the same distribution iff they have the same moment-generating function, so I am trying to show that mgf(u-X) = mgf(X-u)

However, after setting up the integrals I see no way to prove that they are equivalent.

Is this even the correct way to proceed with the problem? Thanks in advance for any suggestions.

  • $\begingroup$ Just find the CDF and show that they are the same function. $$\forall x\in \Bbb R~~\mathsf P(X-\mu\leq x) = \mathsf P(\mu-X\leq x)$$ $\endgroup$ Mar 15, 2017 at 0:05
  • $\begingroup$ Not all symmetric distributions (about their median) have a moment generating function $\endgroup$
    – Henry
    Mar 15, 2017 at 0:33

1 Answer 1


Don't mess around with moment generation, just show directly that the cumulative distributions are the same function.

For all $x\in\Bbb R$ we have to demonstate $\mathsf P(X-\mu\leqslant x)= \mathsf P(\mu-X\leqslant x)$ given that the density function $f$ and constant $\mu$ are such that $\forall x\in\Bbb R~~f(\mu-x)=f(\mu+x)$.


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