Expectation and variance of the poisson distribution

3 Expectation and variance of the poisson distribution

The expectation and variance of the Poisson distribution can be derived directly from the definitions which apply to any discrete probability distribution. However, the algebra involved is a little lengthy. Instead we derive them from the binomial distribution from which the Poisson distribution is derived.

Intuitive Explanation

One way of deriving the mean and variance of the Poisson distribution is to consider the behaviour of the binomial distribution under the following conditions:

$n$ is large
$p$ is small
$n p = λ$ (a constant)

Recalling that the expectation and variance of the binomial distribution are given by the results

$E (X) = n p and V (X) = n p (1 - p) = n p q$

it is reasonable to assert that condition (2) implies, since $q = 1 - p$ , that $q$ is approximately 1 and so the expectation and variance are given by

$E (X) = n p and V (X) = n p q \approx n p$

In fact the algebraic derivation of the expectation and variance of the Poisson distribution shows that these results are in fact exact .

Note that the expectation and the variance are equal.

Key Point 9

The Poisson Distribution

If $X$ is the random variable {number of occurrences in a given interval}

for which the average rate of occurrences is $λ$ and $X$ can assume the values $0, 1, 2, 3, \dots$ and the probability of $r$ occurrences in that interval is given by

$P (X = r) = e^{- λ} \frac{λ^{r}}{r!}$

then the expectation and variance of the distribution are given by the formulae

$E (X) = λ and V (X) = λ$

For a Poisson distribution the Expectation and Variance are equal.

Exercises

Large sheets of metal have faults in random positions but on average have 1 fault per $10 m^{2}$ .
What is the probability that a sheet 5 m $\times$ 8 m will have at most one fault?
If 250 litres of water are known to be polluted with $1 0^{6}$ bacteria what is the probability that a sample of 1 cc of the water contains no bacteria?
Suppose vehicles arrive at a signalised road intersection at an average rate of 360 per hour and the cycle of the traffic lights is set at 40 seconds. In what percentage of cycles will the number of vehicles arriving be
1. exactly 5,
2. less than 5? If, after the lights change to green, there is time to clear only 5 vehicles before the signal changes to red again, what is the probability that waiting vehicles are not cleared in one cycle?
Previous results indicate that 1 in 1000 transistors are defective on average.
1. Find the probability that there are 4 defective transistors in a batch of 2000.
2. What is the largest number, $N$ , of transistors that can be put in a box so that the probability of no defectives is at least 1/2?
A manufacturer sells a certain article in batches of 5000. By agreement with a customer the following method of inspection is adopted: A sample of 100 items is drawn at random from each batch and inspected. If the sample contains 4 or fewer defective items, then the batch is accepted by the customer. If more than 4 defectives are found, every item in the batch is inspected. If inspection costs are 75 p per hundred articles, and the manufacturer normally produces 2 $%$ of defective articles, find the average inspection costs per batch.
A book containing 150 pages has 100 misprints. Find the probability that a particular page contains
1. no misprints,
2. 5 misprints,
3. at least 2 misprints,
4. more than 1 misprint.
For a particular machine, the probability that it will break down within a week is 0.009. The manufacturer has installed 800 machines over a wide area. Calculate the probability that
1. 5,
2. 9,
3. less than 5,
4. more than 4 machines breakdown in a week.
At a given university, the probability that a member of staff is absent on any one day is 0.001. If there are 800 members of staff, calculate the probabilities that the number absent on any one day is
1. 6,
2. 4,
3. 2,
4. 0,
5. less than 3,
6. more than 1.
The number of failures occurring in a machine of a certain type in a year has a Poisson distribution with mean 0.4. In a factory there are ten of these machines. What is
1. the expected total number of failures in the factory in a year?
2. the probability that there are fewer than two failures in the factory in a year?
A factory uses tools of a particular type. From time to time failures in these tools occur and they need to be replaced. The number of such failures in a day has a Poisson distribution with mean 1.25. At the beginning of a particular day there are five replacement tools in stock. A new delivery of replacements will arrive after four days. If all five spares are used before the new delivery arrives then further replacements cannot be made until the delivery arrives.
Find
1. the probability that three replacements are required over the next four days.
2. the expected number of replacements actually made over the next four days.

Poisson Process. In a sheet size $40 m^{2}$ we expect 4 faults
$∴ λ = 4 P (X = r) = λ^{r} e^{- λ} ∕ r!$

$P (X \leq 1) = P (X = 0) + P (X = 1) = e^{- 4} + 4 e^{- 4} = 0.0916$
In 1 cc we expect 4 bacteria $(= 1 0^{6} ∕ 250000) ∴ λ = 4$
$P (X = 0) = e^{- 4} = 0.0183$
In 40 seconds we expect 4 vehicles ∴ λ = 4
1. $P$ (exactly 5) $= λ^{5} e^{- λ} ∕ 5! = 0.15629$ i.e. in 15.6 $%$ of cycles
2. $P$ (less than 5) $= e^{- λ} [1 + λ^{4} + \frac{λ^{2}}{2!} + \frac{λ^{3}}{3!} + \frac{λ^{4}}{4!}]$
$= e^{- 4} [1 + 4 + 8 + \frac{32}{3} + \frac{32}{3}] = 0.6288$

Vehicles will not be cleared if more than 5 are waiting.

$P$ (greater than 5) $= 1 - P$ (exactly 5) $- P$ (less than 5)

$= 1 - 0.15629 - 0.6288 = 0.2148$
1. Poisson approximation to binomial
  $λ = n p = 2000 . \frac{1}{1000} = 2$
  
  $P (X = 4) = λ^{4} e^{- λ} ∕ 4! = 16 e^{- 2} ∕ 24 = 0.09022$
2. $λ = N p = N ∕ 1000; P (X = 0) = \frac{λ^{0} e^{- λ}}{0!} = e^{- λ} = e^{- N ∕ 1000}$
  $e^{- N ∕ 1000} = 0.5 ∴ \frac{- N}{1000} = ln (0.5)$
  
  $∴ N = 693.147 choose N = 693 or less.$

P

(defective)

= 0.02

. Poisson approximation to binomial

λ = n p = 100 (0.02) = 2

$P$ (4 or fewer defectives in sample of 100)

\begin{array}{rcl} = & P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) \\ = & e^{- 2} + 2 e^{- 2} + \frac{2^{2}}{2} e^{- 2} + \frac{2^{3}}{3!} e^{- 2} + \frac{2^{4}}{4!} e^{- 2} = 0.947347 \end{array}

Inspection costs

Cost $c$	75	$75 \times 50$
$P (X = c)$	0.947347	0.0526

$E (Cost) = 75 (0.947347) + 75 \times 50 (0.0526) = 268.5 p$

1. 0.51342
2. 0.00056,
3. 0.14430,
4. 0.14430
1. 0.12038,
2. 0.10698,
3. 0.15552,
4. 0.84448
1. 0.00016,
2. 0.00767,
3. 0.14379,
4. 0.44933,
5. 0.95258,
6. 0.19121
Let X be the total number of failures.
1. $E (X) = 10 \times 0.4 = 4.$
2. $P (X < 2) = P (X = 0) + P (X = 1) = e^{- 4} + 4 e^{- 4} = 5 e^{- 4} = 0.0916 .$
Let the number required over 4 days be X . Then E ( X ) = 4 × 1.25 = 5 and X ∼ Poisson ( 5 ) .
1. $P (X = 3) = \frac{e^{- 5} 5^{3}}{3!} = 0.1404 .$
2. Let $R$ be the number of replacements made.
  
  $E (R) = 0 \times P (X = 0) + \dots + 4 \times P (X = 4) + 5 \times P (X \geq 5),$
  
  and
  
  $P (X \geq 5) = 1 - [P (X = 0) + \dots + P (X = 4)]$
  
  $\begin{array}{rcl} so E (R) & = & 5 - 5 \times P (X = 0) - \dots - 1 \times P (X = 4) \\ = & 5 - e^{- 5} [5 \times \frac{5^{0}}{0!} + 4 \times \frac{5^{1}}{1!} + \dots + 1 \times \frac{5^{4}}{4!}] \\ = & 5 - 0.8773 \\ = & 4.123 . \end{array}$

3 Expectation and variance of the poisson distribution

Key Point 9

Exercises

Answer