91Ó°ÊÓ

X follows a Poisson distribution with parameter \(\lambda \)that is \(X \sim Poisson\left( \lambda \right)\). \(\lambda \) also follows a continuous distribution that is exponential distribution that is \(\lambda \sim \exp \left( 2 \right)\)

We compute first the CDF of\(\lambda \)given X=1.

By hypothesis,

\(\begin{array}{c}P\left( {X = 1|\lambda } \right) = {\rm E}\left( {{1_{X = 1}}|\lambda } \right)\\ = {e^{ - \lambda }}\lambda \end{array}\)

Hence,

\[\begin{aligned}{c}P\left( {X = 1} \right) &= {\rm E}\left( {{1_{X = 1}}} \right)\\ &= {\rm E}\left[ {{\rm E}\left( {{1_{X = 1}}|\lambda } \right)} \right]\\ &= \int\limits_0^\infty {{e^{ - \lambda }}} \lambda f\left( \lambda \right)d\lambda \\ &= \int\limits_0^\infty {2{e^{ - \lambda }}} \lambda {e^{ - 2\lambda }}d\lambda \\ &= \int\limits_0^\infty 2 \lambda {e^{ - 3\lambda }}d\lambda \\ &= \frac{2}{9}\end{aligned}\]

The conditional density function is given by

\({f_{\lambda |X = x}}\left( {t,x} \right) = \frac{{{p_{X = x|\lambda = t}}\left( {x,t} \right){f_\lambda }\left( t \right)}}{{{p_X}\left( x \right)}}\)

Where\({p_{X = x|\lambda = t}}\left( {x,t} \right)\)is the Poisson distribution \(\left( {\lambda = t} \right)\), and \({p_X}\left( x \right)\) is the marginal probability that X=x.found by taking the integral of the numerator with respect to\(\lambda \)from 0 to\(\infty \)

\(\begin{array}{c}{f_{\lambda |X = x}}\left( {t,x} \right) = \frac{{{p_{X = x|\lambda = t}}\left( {x,t} \right){f_\lambda }\left( t \right)}}{{{p_X}\left( x \right)}}\\ = \frac{{\int\limits_0^\lambda {2s{e^{ - 3s}}} ds}}{{P\left( {X = 1} \right)}}\end{array}\)

By differentiating the CDF with respect to\({\bf{\lambda }}\),we obtain the PDF:\(f\left( {\lambda |X = 1} \right) = 9\lambda {e^{ - 3\lambda }}\)

Therefore, theconditionalpdf is\(9\lambda {e^{ - 3\lambda }}\).