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

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii

i

Solution

Multiply the probability density function by

x

and integrate over its support:

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

The middle equality is the beta function integral with parameters

α + 1

and

β

; the last uses the recurrence for the gamma function. Both parameters are positive, so the integral converges. Equivalently, multiplying the beta distribution probability density function by

x

converts its kernel to that of a beta distribution probability density function with its first parameter increased by one.^[ref]

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

to be integers.^[ref]

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