Combining Poisson distributions | CAIE 9709 Probability & Statistics 2 Notes

Chapters

Probability & Statistics 2

Mathematics9709

Only independent sums keep the Poisson model

Which expressions stay Poisson?

X + Y

Po(λ₁ + λ₂)

YES — if independent

X − Y or 2X − Y

calculate E and Var only

NO — not Poisson

Equal mean and variance is necessary for Poisson, but not sufficient.

Poisson closure is narrow: independent sums only.

X+Y\sim\operatorname{Po}(\lambda_X+\lambda_Y)

Symbols

$X,Y$ = independent Poisson variables (counts)
$\lambda_X+\lambda_Y$ = combined mean rate (events)

Worked example

Current-paper combined Poisson count

$X\sim\operatorname{Po}(3)$ and $Y\sim\operatorname{Po}(2)$ are independent. Find $P(X+Y>2)$ .

T=X+Y\sim\operatorname{Po}(5)

P(T>2)=1-P(T\le2)

Answer

P(T>2)=0.875

(9709/61/O/N/25 Q1(b))

A difference or multiple usually is not Poisson

Worked example

Calculate mean and variance without naming a distribution

$X\sim\operatorname{Po}(2.4)$ and $Y\sim\operatorname{Po}(3.6)$ are independent. Let $T=2X-Y$ .

E(T)=2(2.4)-3.6=1.2

2^2(2.4)+3.6=13.2

T is not Poisson. Its mean and variance are unequal, and the subtraction can also produce negative values.

Answer

E(T)=1.2

\operatorname{Var}(T)=13.2

Key idea

Equal mean and variance is necessary for a Poisson distribution, but it is not enough by itself to prove that a variable is Poisson.

Decide the distribution before finding probability

Check that every component is Poisson.
Check independence.
Check that the operation is a sum with coefficient 1.
If any check fails, use general mean and variance rules only.

Your turnIndependent Poisson-combination check [4 marks]

A, B and C are independent with means 2, 1.2 and 0.5 and each has a Poisson distribution. Find $P(A+B+C\le2)$ .

Show worked answer

T=A+B+C\sim\operatorname{Po}(3.7)

Answer

P(T\le2)=0.285

Common mistake

Do not call

X-Y

Poisson. A Poisson count cannot be negative, but a difference of counts can be.

Only independent sums keep the Poisson model

Which expressions stay Poisson?

X + Y

Po(λ₁ + λ₂)

YES — if independent

X − Y or 2X − Y

calculate E and Var only

NO — not Poisson

Equal mean and variance is necessary for Poisson, but not sufficient.

Poisson closure is narrow: independent sums only.

X+Y\sim\operatorname{Po}(\lambda_X+\lambda_Y)

Symbols

$X,Y$ = independent Poisson variables (counts)
$\lambda_X+\lambda_Y$ = combined mean rate (events)

Worked example

Current-paper combined Poisson count

$X\sim\operatorname{Po}(3)$ and $Y\sim\operatorname{Po}(2)$ are independent. Find $P(X+Y>2)$ .

T=X+Y\sim\operatorname{Po}(5)

P(T>2)=1-P(T\le2)

Answer

P(T>2)=0.875

(9709/61/O/N/25 Q1(b))

A difference or multiple usually is not Poisson

Worked example

Calculate mean and variance without naming a distribution

$X\sim\operatorname{Po}(2.4)$ and $Y\sim\operatorname{Po}(3.6)$ are independent. Let $T=2X-Y$ .

E(T)=2(2.4)-3.6=1.2

2^2(2.4)+3.6=13.2

T is not Poisson. Its mean and variance are unequal, and the subtraction can also produce negative values.

Answer

E(T)=1.2

\operatorname{Var}(T)=13.2

Key idea

Equal mean and variance is necessary for a Poisson distribution, but it is not enough by itself to prove that a variable is Poisson.

Decide the distribution before finding probability

Check that every component is Poisson.
Check independence.
Check that the operation is a sum with coefficient 1.
If any check fails, use general mean and variance rules only.

Your turnIndependent Poisson-combination check [4 marks]

A, B and C are independent with means 2, 1.2 and 0.5 and each has a Poisson distribution. Find $P(A+B+C\le2)$ .

Show worked answer

T=A+B+C\sim\operatorname{Po}(3.7)

Answer

P(T\le2)=0.285

Common mistake

Do not call

X-Y

Poisson. A Poisson count cannot be negative, but a difference of counts can be.