3
$\begingroup$

I would like to 'prove' that
$$q_{\alpha}(X + c) = q_{\alpha}(X) + c $$ For c $\in \Bbb{R}$, $X$ a random variable, and $q_{\alpha}$ the quantile of order ${\alpha}$. I would actually like to prove this for conditional quantile (but I think it does not really impact the proof).

I saw the 'equivariance of quantile under monotone transformation' property, which is a way stronger property (and I did not find any proof for this either anyway), so I think that $q_{\alpha}(X + c) = q_{\alpha}(X) + c $ is true (or maybe there are some particular cases in which it does not hold? I cannot think of any). The only problem is I don't know how to prove it.
To me it seems quite intuitive, but I'm not sure I am allowed to write this without proving it (and there might be some special case I'm not thinking of).

I would also appreciate if anyone got a link to a formal proof of the equivariance of quantile under monotone transformation.

Thanks in advance.

$\endgroup$
2
$\begingroup$

This is based on $$ P(X+a > t) = P(X > t - a) $$ connected to the definition of $q_\alpha$.

$\endgroup$
  • $\begingroup$ I thought there was something more tricky (i.e. obliged to use the Inf { x: F(x) >= alpha } definition for special cases), but anyway, this seem to work , thanks! $\endgroup$ – G. Ander Oct 29 '15 at 14:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.