91Ó°ÊÓ

The rate β of the Poisson process is unknown and has the gamma distribution with parameters α and γ. Let X be the number of particles that strike the target during t time units. We need to prove that the conditional distribution of β given X = x is a gamma distribution, and find the parameters of that gamma distribution.

The conditional PDF of X given β is Poisson distribution

The distribution of β hasthe gamma distribution with parameters α and γ.

The conditional PDF of β is

\({f_1}\left( {x|\beta } \right) = \frac{{{e^{ - \beta t}}}}{{x!}}{\left( {\beta t} \right)^x}\)and marginal PDF of β is\({f_2}\left( \beta \right) = \frac{{{\gamma ^\alpha }}}{{\left| \!{\overline {\, \alpha \,}} \right. }}{\beta ^{x + \alpha - 1}}{e^{ - \beta \gamma }}\)

So, the joint PDF of X and β is

\(\begin{aligned}{}f\left( {x,\beta } \right)& = {f_1}\left( {x|\beta } \right){f_2}\left( \beta \right)\\& = \frac{{{e^{ - \beta t}}{{\left( {\beta t} \right)}^x}}}{{x!}} \times \frac{{{\gamma ^\alpha }}}{{\left| \!{\overline {\, \alpha \,}} \right. }}{\beta ^{x + \alpha - 1}}{e^{ - \beta \gamma }}\\ &= \frac{{{\gamma ^\alpha }{t^x}}}{{\left| \!{\overline {\, \alpha \,}} \right. \times x!}}{\beta ^{x + \alpha - 1}}{e^{ - \beta \left( {t + \alpha } \right)}}\\\end{aligned}\)

Again the marginal distribution of X is

\(\begin{aligned}{}{f_3}\left( x \right) &= \int {f\left( {x,\beta } \right)d\beta } \\ &= \int\limits_0^\infty {\frac{{{e^{ - \beta t}}{{\left( {\beta t} \right)}^x}}}{{x!}} \times \frac{{{\gamma ^\alpha }}}{{\left| \!{\overline {\, \alpha \,}} \right. }}{\beta ^{x + \alpha - 1}}{e^{ - \beta \gamma }}\,d\beta } \\& = \frac{{{\gamma ^\alpha}{t^x}}}{{\left| \!{\overline {\, \alpha \,}} \right.!}}\int\limits_0^\infty {{\beta ^{x + \alpha - 1}}{e^{ - \beta \left( {t +\gamma } \right)}}} d\beta \\& = \frac{{{\gamma ^\alpha }{t^x}}}{{\left| \!{\overline{\, \alpha \,}} \right. \times x!}} \times \frac{{\left| \!{\overline {\, {\left( {x + \alpha } \right)} \,}} \right. }}{{{{\left( {t + \gamma } \right)}^{\left( {x + \alpha } \right)}}}}\end{aligned}\)

Now, the conditional PDF of β given X = x is \(\lambda = t + \gamma \,{\rm{and}}\,r = x + \alpha \)\(\begin{aligned}{}{f_4}\left( {\beta |x} \right) &= \frac{{f\left( {x,\beta } \right)}}{{{f_3}\left( x \right)}}\\& = \frac{{\frac{{{\gamma^\alpha}{t^x}}}{{\left|\!{\overline {\, \alpha \,}} \right. \times x!}}{\beta ^{x + \alpha - 1}}{e^{ - \beta \left( {t + \alpha } \right)}}}}{{\frac{{{\gamma ^\alpha }{t^x}}}{{\left| \!{\overline {\, \alpha \,}} \right. \times x!}} \times \frac{{\left| \!{\overline {\, {\left( {x + \alpha } \right)} \,}} \right. }}{{{{\left( {t + \gamma } \right)}^{\left( {x + \alpha } \right)}}}}}}\\ &= \frac{{{{\left( {t + \gamma } \right)}^{\left( {x + \alpha } \right)}}}}{{\left| \!{\overline {\, {\left( {x + \alpha } \right)} \,}} \right. }}{\beta ^{x + \alpha - 1}}{e^{ - \beta \left( {t + \alpha } \right)}}\end{aligned}\)

This is clearly gamma distribution with parameter \(\lambda = t + \gamma \,{\rm{and}}\,r = x + \alpha \).