Last updated: 2019-05-19

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Rmd 15aa416 Jason Willwerscheid 2019-05-19 wflow_publish(“analysis/poisson_p.Rmd”)

Consider \[ X_i \sim \text{Poisson}(\lambda_i) \] No matter the distribution of \(\lambda\), \(p\)-values won’t be uniformly distributed. To see this, note that \[ P(p_i < \alpha) = P(X_i \le \text{qpois}(\alpha, \lambda_i) - 1) = \text{ppois}(\text{qpois}(\alpha, \lambda_i) - 1, \lambda_i)\]

Setting \(\alpha = 0.05\):

lambda <- seq(0.01, 10, by = .01)
df <- data.frame(lambda = lambda, y = ppois((qpois(.05, lambda) - 1), lambda))
ggplot(df, aes(x = lambda, y = y)) + geom_point() +
  labs(y = expression(P(p < 0.05)))

So \(P(p < .05)\) is guaranteed to be less than .05, especially for distributions of \(\lambda\) that have a large mass on small values of \(\lambda\).

Similarly, \[ P(p_i > \alpha) = P(X_i \ge \text{qpois}(\alpha, \lambda_i)) = 1 - P(X_i \le \text{qpois}(\alpha, \lambda_i) - 1)\]

For \(\alpha = .95\):

df <- data.frame(lambda = lambda, y = 1 - ppois(qpois(.95, lambda) - 1, lambda))
ggplot(df, aes(x = lambda, y = y)) + geom_point() +
  labs(y = expression(P(p > 0.95))) + geom_abline(slope = 0, intercept = .05)

So \(P(p > .95)\) is guaranteed to be greater than .05. 

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