testing by betting—simple vs simple

hypothesis-testingtesting-by-bettingsequential-statisticslikelihood-ratio

Here we explore game-theoretic hypothesis testing for simple nulls vs simple alternatives. Suppose we observe random variables $X_1,X_2,\dots$ and are testing

$$H_0:(X_t)\sim P{H}_1:(X_t)\sim Q,$$

for two distributions $P$ and $Q$. We want to design a payoff function $S_t$ such that

$${\mathbb{E}}_P[S_t(X_t)|{\mathcal{F}}_{t-1}]=1.$$

Assuming densities of $P$ and $Q$ are well-defined (otherwise we can use Radon-Nikodym derivatives), note that this implies

$$1={\mathbb{E}}_P[S_t(X_t)|{\mathcal{F}}_{t-1}]=\int S_t(x)p(x|{\mathcal{F}}_{t-1})dx,$$

so $S_t(x)p(x|{\mathcal{F}}_{t-1})$ is a density, call it $r(x|{\mathcal{F}}_{t-1})$. Hence $S_t(x)=r(x|{\mathcal{F}}_{t-1})/p(x|{\mathcal{F}}_{t-1})$. So in order to maximize log wealth (see the principle of maximizing log-wealth), we want $r$ such that

$$\mathbb{E}\left[\log \left(\frac{r(x)}{p(x)}\right)|{\mathcal{F}}_{t-1}\right]\ge \mathbb{E}\left[\log \left(\frac{u(x)}{p(x)}\right)|{\mathcal{F}}_{t-1}\right],$$

for all distributions $u$. It turns out that by Gibbs’ inequality, the optimal distribution to pick is simply $Q$. Therefore, our bet is the likelihood ratio

$$S_t(X_t)=\frac{q(X_t|{\mathcal{F}}_{t-1})}{p(X_t|{\mathcal{F}}_{t-1})}.$$

The corresponding wealth process $K_t={\prod }_{i\le t}{S}_i(X_i)=q(X_1^t)/p(X_1^t)$ therefore recovers the classical likelihood-ratio test (and is a sequentialized version of it, the sequential probability ratio test).

Note that this shows that the GRO (growth rate optimal) (see growth rate conditions in sequential testing) e-value for simple vs simple hypothesis testing is the likelihood ratio.

For composite nulls and/or alternatives see testing by betting—simple vs composite or testing by betting—composite vs composite.