# Show that $F_X(q_t)=t$

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 $$F_X(q_t)=t.$$

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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 $F_X(a)=t$.