1.4 Computing Probabilities by Conditioning
Let \(E\) be an event and \(I_E\) its indicator. Then \(E(I_E) = P(E)\), and \(E(I_E\mid Y = y) = P(E\mid Y = y)\). Applying lemma 1.3.1 to \(I_E\) gives \[P(E) = \sum _yP(E\mid Y = y)P(Y = y)\qquad \text {or}\qquad P(E) = \int _{-\infty }^\infty P(E\mid Y = y)f_Y(y)\,dy\] for discrete and continuous \(Y\) respectively: the law of total probability.
Example 1.4.1 (Accident proneness). An insurance company models the number of accidents a policy holder has in a year as Poisson, with a mean \(\Lambda \) that varies between policy holders. Suppose \(\Lambda \) has the gamma density \(g(\lambda ) = \lambda e^{-\lambda }\), \(\lambda \geq 0\). Find the probability that a randomly chosen policy holder has exactly \(n\) accidents in a year.
Solution. Conditioning on \(\Lambda = \lambda \), \[P(X = n) = \int _0^\infty \frac {e^{-\lambda }\lambda ^n}{n!}\,\lambda e^{-\lambda }\,d\lambda = \frac {1}{n!}\int _0^\infty \lambda ^{n + 1}e^{-2\lambda }\,d\lambda .\] With \(m = 2\lambda \) the integral is \(\frac {1}{2^{n + 2}}\int _0^\infty m^{n + 1}e^{-m}\,dm = \frac {\Gamma (n + 2)}{2^{n + 2}} = \frac {(n + 1)!}{2^{n + 2}}\), so \[P(X = n) = \frac {n + 1}{2^{n + 2}},\qquad n = 0, 1, 2, \dots .\] These sum to \(1\), as they must, and form a negative binomial distribution. Mixing Poisson distributions over a gamma mean always gives a negative binomial; its variance exceeds its mean, which is why accident and claim counts are often over-dispersed relative to the Poisson.
Example 1.4.2. \(X\) and \(Y\) are independent, \(X\) exponential with rate \(\lambda \) and \(Y\) exponential with rate \(\mu \). Find \(P(X<Y)\).
Solution. Conditioning on \(Y = y\), \(P(X<Y\mid Y = y) = P(X<y) = 1 - e^{-\lambda y}\), so \[P(X<Y) = \int _0^\infty (1 - e^{-\lambda y})\mu e^{-\mu y}\,dy = 1 - \frac {\mu }{\lambda + \mu } = \frac {\lambda }{\lambda + \mu } .\] This result is used repeatedly in continuous-time chains: of two competing exponential clocks, the one with rate \(\lambda \) rings first with probability \(\frac {\lambda }{\lambda + \mu }\).
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.