1
$\begingroup$

I am studying for an exam in actuarial science, where I have the following exercise:

Prove that the stochastic order relation $\leq_{\mathrm{st}}$ is functionally invariant; i.e. show that $$X \leq_{\mathrm{st}} Y \implies f(X) \leq_{\mathrm{st}}f(Y)$$ for any non-decreasing function $f(\cdot)$.

I suspect I need to use the Theorem $$X \leq_{\mathrm{st}} Y \iff F_X(x) \geq F_Y(x) \quad \text{for all } x,$$ where $F_X$ is the cumulative distribution function of $X$, but I don't know how to continue. Any suggestions?

$\endgroup$
1
$\begingroup$

I'm not sure which stochastic ordering you are working with so I will use the theorem to prove the statement: $$X \leq_{\mathrm{st}} Y \iff F_X(x) \geq F_Y(x) \quad \text{for all } x,$$ $$P(X \leq x) \leq P(Y \leq x)$$

So we want to show that: $$P(f(X) \leq x) \leq P(f(Y) \leq x)$$ i.e. $$P(f(X) \in (-\infty, x]) \leq P(f(Y) \in (-\infty, x])$$

Suppose $f$ is left-continuous and let $y$ be the maximum value s.t. $f(y) \leq x$. Since $f$ is left-continuous and non-decreasing a specific value always exists and $$f(X) \in (-\infty, x] \iff X \in (-\infty, y]$$ Therefore, $$P(X \in (-\infty, y]) \leq P(Y \in (-\infty, y])$$ is always true by supposition. You will now need to handle those locations where $f$ is not left-continuous.

$\endgroup$

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.