Proving Poisson properties from first principles

A question is this type if and only if it requires algebraic proof of Poisson distribution properties such as E(X), Var(X), or E(X(X-1)) using the probability mass function.

7 questions · Standard +0.8

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AQA S3 2008 June Q7
19 marks Standard +0.8
7
  1. The random variable \(X\) has a Poisson distribution with \(\mathrm { E } ( X ) = \lambda\).
    1. Prove, from first principles, that \(\mathrm { E } ( X ( X - 1 ) ) = \lambda ^ { 2 }\).
    2. Hence deduce that \(\operatorname { Var } ( X ) = \lambda\).
  2. The independent Poisson random variables \(X _ { 1 }\) and \(X _ { 2 }\) are such that \(\mathrm { E } \left( X _ { 1 } \right) = 5\) and \(\mathrm { E } \left( X _ { 2 } \right) = 2\). The random variables \(D\) and \(F\) are defined by $$D = X _ { 1 } - X _ { 2 } \quad \text { and } \quad F = 2 X _ { 1 } + 10$$
    1. Determine the mean and the variance of \(D\).
    2. Determine the mean and the variance of \(F\).
    3. For each of the variables \(D\) and \(F\), give a reason why the distribution is not Poisson.
  3. The daily number of black printer cartridges sold by a shop may be modelled by a Poisson distribution with a mean of 5 . Independently, the daily number of colour printer cartridges sold by the same shop may be modelled by a Poisson distribution with a mean of 2. Use a distributional approximation to estimate the probability that the total number of black and colour printer cartridges sold by the shop during a 4 -week period ( 24 days) exceeds 175.
AQA S3 2010 June Q7
15 marks Challenging +1.2
7 The random variable \(X\) has a Poisson distribution with parameter \(\lambda\).
    1. Prove, from first principles, that \(\mathrm { E } ( X ) = \lambda\).
    2. Hence, given that \(\mathrm { E } ( X ( X - 1 ) ) = \lambda ^ { 2 }\), find, in terms of \(\lambda\), an expression for \(\operatorname { Var } ( X )\).
  2. The mode, \(m\), of \(X\) is such that $$\mathrm { P } ( X = m ) \geqslant \mathrm { P } ( X = m - 1 ) \quad \text { and } \quad \mathrm { P } ( X = m ) \geqslant \mathrm { P } ( X = m + 1 )$$
    1. Show that \(\lambda - 1 \leqslant m \leqslant \lambda\).
    2. Given that \(\lambda = 4.9\), determine \(\mathrm { P } ( X = m )\).
  3. The random variable \(Y\) has a Poisson distribution with mode \(d\) and standard deviation 15.5. Use a distributional approximation to estimate \(\mathrm { P } ( Y \geqslant d )\).
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AQA S3 2011 June Q7
9 marks Standard +0.8
7
  1. The random variable \(X\) has a Poisson distribution with \(\mathrm { E } ( X ) = \lambda\).
    1. Prove, from first principles, that \(\mathrm { E } ( X ( X - 1 ) ) = \lambda ^ { 2 }\).
    2. Hence deduce that \(\operatorname { Var } ( X ) = \mathrm { E } ( X )\).
  2. The random variable \(Y\) has a Poisson distribution with \(\mathrm { E } ( Y ) = 2.5\). Given that \(Z = 4 Y + 30\) :
    1. show that \(\operatorname { Var } ( Z ) = \mathrm { E } ( Z )\);
    2. give a reason why the distribution of \(Z\) is not Poisson.
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AQA S3 2006 June Q6
8 marks Challenging +1.2
6 The random variable \(X\) has a Poisson distribution with parameter \(\lambda\).
  1. Prove that \(\mathrm { E } ( X ) = \lambda\).
  2. By first proving that \(\mathrm { E } ( X ( X - 1 ) ) = \lambda ^ { 2 }\), or otherwise, prove that \(\operatorname { Var } ( X ) = \lambda\).
Pre-U Pre-U 9795/2 2015 June Q4
11 marks Challenging +1.2
4
  1. (a) Derive the moment generating function for a Poisson distribution with mean \(\lambda\).
    (b) The independent random variables \(X\) and \(Y\) are such that \(X \sim \operatorname { Po } ( \mu )\) and \(Y \sim \operatorname { Po } ( v )\). Use moment generating functions to show that \(( X + Y ) \sim \operatorname { Po } ( \mu + v )\).
  2. The number of goals scored per match by Camford Academicals FC may be modelled by a Poisson distribution with mean 2. The number of goals scored against Camford during a match may be modelled by an independent Poisson distribution with mean \(k\). The probability that no goals are scored, by either side, in a match involving Camford is 0.045 . Find
    (a) the value of \(k\),
    (b) the probability that exactly 3 goals are scored against Camford in a match,
    (c) the probability that the total number of goals scored, in a match involving Camford, is between 2 and 5 inclusive.
AQA S3 2016 June Q6
22 marks Standard +0.3
  1. The discrete random variable \(X\) has probability distribution given by $$\mathrm{P}(X = x) = \begin{cases} \frac{e^{-\lambda}\lambda^x}{x!} & x = 0, 1, 2, \ldots \\ 0 & \text{otherwise} \end{cases}$$ Show that \(\mathrm{E}(X) = \lambda\) and that \(\mathrm{Var}(X) = \lambda\). [7 marks]
  2. In light-weight chain, faults occur randomly and independently, and at a constant average rate of 0.075 per metre.
    1. Calculate the probability that there are no faults in a 10-metre length of this chain. [2 marks]
    2. Use a distributional approximation to estimate the probability that, in a 500-metre reel of light-weight chain, there are:
      1. fewer than 30 faults;
      2. at least 35 faults but at most 45 faults.
      [7 marks]
  3. As part of an investigation into the quality of a new design of medium-weight chain, a sample of fifty 10-metre lengths was selected. Subsequent analysis revealed a total of 49 faults. Assuming that faults occur randomly and independently, and at a constant average rate, construct an approximate 98\% confidence interval for the average number of faults per metre. [6 marks]
Pre-U Pre-U 9795/2 2014 June Q5
13 marks Standard +0.3
  1. The discrete random variable \(X\) has a Poisson distribution with mean \(\lambda\). Use the probability generating function for \(X\) to show that both the mean and the variance have the value \(\lambda\). [5]
  2. The number of eggs laid by a certain insect has a Poisson distribution with variance 250. Find, using a suitable approximation, the probability that between 230 and 260 (inclusive) eggs are laid. [5]
  3. An insect lays 250 eggs. The probability that any egg that is laid survives to maturity is 0.1. Use a suitable approximation to find the probability that more than 30 eggs survive to maturity. [3]