 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2022/discrete-mathematics-continuous-mathematics-and-probability-resit/9/b/ii/solution

The same integral works with both powers shifted:

E [X^{s} (1 - X)^{t}] = \frac{Γ ( α + β )}{Γ ( α ) Γ ( β )} \int_{0}^{1} x^{α + s - 1} (1 - x)^{β + t - 1} d x = \frac{Γ ( α + β ) Γ ( α + s ) Γ ( β + t )}{Γ ( α ) Γ ( β ) Γ ( α + β + s + t )} .

The assumptions

s > - α

and

t > - β

say precisely that the exponents at each endpoint are greater than

- 1

, ensuring integrability there. Values of a negative power at the endpoints themselves can be assigned arbitrarily, because those endpoints have probability zero. This establishes the mixed power moments of a beta distribution without requiring

s

and

t

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