First-order stochastic dominance and truncation

Question

Suppose we have two distributions $F$ and $G$ over $\left[0,1\right]$. Suppose $F(x) \leq G(x)$ for all $x$, i.e. $F$ first-order stochastically dominates $G$. Is it true that $F(x|x\leq k) \leq G(x|x\leq k)$ for all $k$ and for all $x \in \left[0,k\right]$? Put in another way, does first-order stochastic dominance survive half-truncations?

Thanks.

Answer 1

Lets define a the truncated distribution functions as:

$F_k(x)=a_kF(x)\mathbf{1}_{\leq k}(x)$ and $G_k(x)=b_kG(x)\mathbf{1}_{\leq k}(x)$ for $0 <b_k=\frac{1}{G(k)}\leq a_k=\frac{1}{F(k)}$.

Assume Condition 1: $\exists(k\in [0,1], c_k\in [0,k]): \frac{G(c_k)}{F(c_k)}<\frac{G(k)}{F(k)}$.

Then, $\frac{b_kG(c_k)}{a_kF(c_k)}=\frac{F(k)G(c_k)}{G(k)F(c_k)}<1$

Since we are talking about arbitrary distributions, first order stochastic dominance is not stable under half-truncations, as Condition 1 does not preclude $F$ from unconditionally dominating $G$.