Raising the bar

Analysis of problems with conditioning

These problems are among the most difficult. It's important to work out a general approach to such problems. All references are to J. Abdey,  Advanced statistics: distribution theory, ST2133, University of London, 2021.

General scheme

Step 1. Conditioning is usually suggested by the problem statement: $latex Y$ is conditioned on $latex X$.

Your life will be easier if you follow the notation used in the guide: use $latex p $ for probability mass functions (discrete variables) and $latex f$ for (probability) density functions (continuous variables).

a) If $latex Y|X$ and $latex X$ both are discrete (Example 5.1, Example 5.13, Example 5.18):

$latex p_{Y}\left( y\right) =\sum_{Set}p_{Y\vert X}\left( y\vert x\right) p_{X}\left(x\right) .$

b) If $latex Y|X$ and $latex X$ both are continuous (Activity 5.6):

$latex f_{Y}\left( y\right) =\int_{Set}f_{Y\vert X}\left( y\vert x\right) f_{X}\left(x\right) dx.$

c) If $latex Y|X$ is discrete, $latex X$ is continuous (Example 5.2, Activity 5.5):

$latex p_{Y}\left( y\right) =\int_{Set}p_{Y\vert X}\left( y\vert x\right) f_{X}\left(x\right) dx$

d) If $latex Y|X$ is continuous, $latex X$ is discrete (Activity 5.12):

$latex f_{Y}\left( y\right) =\sum_{Set}f_{Y\vert X}\left( y\vert x\right) p_{X}\left(x\right) .$

In all cases you need to figure out $latex Set$ over which to sum or integrate.

Step 2. Write out the conditional densities/probabilities with the same arguments
as in your conditional equation.

Step 3. Reduce the result to one of known distributions using the completeness
axiom.

Example 5.1

Let $latex X$ denote the number of hurricanes which form in a given year, and let $latex Y$ denote the number of these which make landfall. Suppose each hurricane has a probability of $latex \pi $ making landfall independent of other hurricanes. Given the number of hurricanes $latex x$, then $latex Y$ can be thought of as the number of successes in $latex x$ independent and identically distributed Bernoulli trials. We can write this as $latex Y|X=x\sim Bin(x,\pi )$. Suppose we also have that $latex X\sim Pois(\lambda )$. Find the distribution of $latex Y$ (noting that $latex X\geq Y$ ).

Solution

Step 1. The number of hurricanes $latex X$ takes values $latex 0,1,2,...$ and is distributed as Poisson. The number of landfalls for a given $latex X=x$ is binomial with values $latex y=0,...,x$. It follows that $latex Set={x:x\ge y}$.

Write the general formula for conditional probability:

$latex p_{Y}\left( y\right) =\sum_{x=y}^{\infty }p_{Y\vert X}\left( y\vert x\right)p_{X}\left( x\right) .$

Step 2. Specifying the distributions:

$latex p_{X}\left( x\right) =e^{-\mu }\frac{\mu ^{x}}{x!},$ where $latex x=0,1,2,...,$

and

$latex P\left( Bin\left( x,\pi \right) =y\right) =p_{Y\vert X}\left( y\vert x\right)=C_{x}^{y}\pi ^{y}\left( 1-\pi \right) ^{x-y}$ where $latex y\leq x.$

Step 3. Reduce the result to one of known distributions:

$latex p_{Y}\left( y\right) =\sum_{x=y}^{\infty }C_{x}^{y}\pi ^{y}\left( 1-\pi\right) ^{x-y}e^{-\mu }\frac{\mu ^{x}}{x!}$

(pull out of summation everything that does not depend on summation variable
$latex x$)

$latex =\frac{e^{-\mu }\mu ^{y}}{y!}\pi ^{y}\sum_{x=y}^{\infty }\frac{1}{\left(x-y\right) !}\left( \mu \left( 1-\pi \right) \right) ^{x-y}$

(replace $latex x-y=z$ to better see the structure)

$latex =\frac{e^{-\mu }\mu ^{y}}{y!}\pi ^{y}\sum_{z=0}^{\infty }\frac{1}{z!}\left(\mu \left( 1-\pi \right) \right) ^{z}$

(using the completeness axiom $latex \sum_{x=0}^{\infty }\frac{\mu ^{x}}{x!}=e^{\mu }$ for the Poisson variable)

$latex =\frac{e^{-\mu }}{y!}\left( \mu \pi \right) ^{y}e^{\mu \left( 1-\pi \right)}=\frac{e^{-\mu \pi }}{y!}\left( \mu \pi \right) ^{y}=p_{Pois(\mu \pi)}\left( y\right) .$