Home page for accesible maths 6 Expectation (II)6.2 Conditional Expectations 6.4 Moment generating functions

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6.3 Decomposition of the marginal variance

We have seen that the marginal expectations can be obtained from the conditional expectations. We can also obtain the marginal variances from the conditional expectations and variances by the following formula:

	$\displaystyle\operatorname{\mathsf{E}}\left[{{\operatorname{\mathsf{Var}}}% \left[{Y\mid X}\right]}\right]+{\operatorname{\mathsf{Var}}}\left[{% \operatorname{\mathsf{E}}\left[{Y\mid X}\right]}\right]$
	$\displaystyle=\operatorname{\mathsf{E}}\left[{\operatorname{\mathsf{E}}\left[{% Y^{2}\mid X}\right]-\operatorname{\mathsf{E}}\left[{Y\mid X}\right]^{2}}\right% ]+\operatorname{\mathsf{E}}\left[{\operatorname{\mathsf{E}}\left[{Y\mid X}% \right]^{2}}\right]-\operatorname{\mathsf{E}}\left[{\operatorname{\mathsf{E}}% \left[{Y\mid X}\right]}\right]^{2}$
	$\displaystyle=\operatorname{\mathsf{E}}\left[{Y^{2}}\right]-\operatorname{% \mathsf{E}}\left[{Y}\right]^{2}$
	$\displaystyle={\operatorname{\mathsf{Var}}}\left[{Y}\right].$

These formulae are particularly useful when a random variable $Y$ is given as a mixture of distributions. This is most easily illustrated by an example.

Example 6.3.1.

Let $X$ be a Poisson $(\lambda)$ random variable, and given $X$ takes the value $x$ let $Y$ be Binomial $(x,p)$ -distributed, i.e. $Y\mid X=x\sim$ Binomial $(x,p)$ . Find the expectation and variance of $Y$ .

Solution. From properties of the Binomial distribution we have

$\operatorname{\mathsf{E}}\left[{Y\mid X=x}\right]=xp$ ,
${\operatorname{\mathsf{Var}}}\left[{Y\mid X=x}\right]=xp(1-p)$ .

Hence, using properties of the Poisson distribution we obtain

\displaystyle\operatorname{\mathsf{E}}\left[{Y}\right]=\operatorname{\mathsf{E% }}\left[{\operatorname{\mathsf{E}}\left[{Y\mid X}\right]}\right]=\operatorname% {\mathsf{E}}\left[{Xp}\right]=\lambda p.

	$\displaystyle{\operatorname{\mathsf{Var}}}\left[{Y}\right]$	$\displaystyle=\operatorname{\mathsf{E}}\left[{{\operatorname{\mathsf{Var}}}% \left[{Y\mid X}\right]}\right]+{\operatorname{\mathsf{Var}}}\left[{% \operatorname{\mathsf{E}}\left[{Y\mid X}\right]}\right]$
		$\displaystyle=\operatorname{\mathsf{E}}\left[{Xp(1-p)}\right]+{\operatorname{% \mathsf{Var}}}\left[{Xp}\right]$
		$\displaystyle=\lambda p(1-p)+\lambda p^{2}$
		$\displaystyle=\lambda p.$

In fact, it can be shown that $Y\sim\operatorname{Poisson}(\lambda p)$ .