You shouldn’t need to do any integrals to show that the PIT gives a uniform distribution. Suppose Pr(X ⇐ x) = p; then (assuming no jumps in the cdf) the PIT maps x to p. In other words, writing P for the random variable produced by the PIT, Pr(P ⇐ p) = p, so P is uniform.
Yup, that works. I would only caution that “assuming no jumps in the cdf” is not quite the right condition: singular distributions (e.g., the Cantor distribution) contain jumps, and the PIT works fine for them. The correct condition is that the random variable not have a discrete component.
You shouldn’t need to do any integrals to show that the PIT gives a uniform distribution. Suppose Pr(X ⇐ x) = p; then (assuming no jumps in the cdf) the PIT maps x to p. In other words, writing P for the random variable produced by the PIT, Pr(P ⇐ p) = p, so P is uniform.
Yup, that works. I would only caution that “assuming no jumps in the cdf” is not quite the right condition: singular distributions (e.g., the Cantor distribution) contain jumps, and the PIT works fine for them. The correct condition is that the random variable not have a discrete component.
Sure.
Pr(X ⇐ x) is an integral, but I find this explanation clearer than the one in the OP. Upvoted.