# Is it true that $X(t)^a > K \iff X(t) > K^\frac1a$

Let $a \in \mathbb{N}$, $K \in \mathbb{R^+}$ and $X(t)$ be a geometric Brownian Motion. Is the following true?

$$X(t)^a > K \iff X(t) > K^\frac1a$$

The context of the above is that I want to evaluate the probability that $X(t)^a > K$, and this transformation of the inequality would be helpful.

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Is $X(t)\ge 0$ also guaranteed? –  Berci Nov 23 '12 at 12:21
@Berci Yes. $\phantom{.}$ –  Jase Nov 23 '12 at 12:26
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## 1 Answer

Yes. The function $\mathbb R^+_0 \to \mathbb R^+_0$, $x \mapsto x^{1/a}$ is strictly monotone.

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