Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a scalar random variable having $F_X$ as its cdf. Let $p_t$ be the function defined by $p_t(u)=u(t-\mathbf 1_{u\lt0})$ where $\mathbf 1$ denotes the indicator function. Let $$q_t=\arg\min E\left[p_t(X-a)\right].$$

Show that $$F_X(q_t)=t.$$

share|cite|improve this question
up vote 1 down vote accepted

Let $u(a)=\mathrm E(p_t(X-a))$. Since $p_t(x-a)=tx-ta+(a-x)^+=tx-ta+\int\limits_{-\infty}^a\mathbf 1_{x\leqslant b}\mathrm db$, $$ u(a)=t\mathrm E(X)-ta+\int\limits_{-\infty}^aF_X(b)\mathrm db,\qquad u'(a)=-t+F_X(a). $$ This shows that $u$ is decreasing at $a$ if $F_X(a)\lt t$ and increasing at $a$ if $F_X(a)\gt t$. Hence the minimum of $u$ is reached at every $a$ such that $F_X(a)=t$.

share|cite|improve this answer

Your Answer


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

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