Coursework solutions for Math 103 Probability: Week 13

1. This can be done in (at least) two ways (starting from either $\mathrm{Var}(R)=\mathrm{E}[{(R-\mathrm{E}[R])}^2]$ or as below). Working from the alternative formulation of variance gives the easiest route:

   $\mathrm{Var}(R)$ $=$ $\mathrm{E}[R^2]-{(\mathrm{E}[R])}^2$

   $=$ $\mathrm{E}[R(R-1)+R]-{(\mathrm{E}[R])}^2$

   $=$ $\mathrm{E}[R(R-1)]+\mathrm{E}[R]-{(\mathrm{E}[R])}^2$

   Marks awarded as follows: start at 3 marks; deduct 1 if the start point is not clearly specified; deduct 1 if there is a major gap in the logic; deduct 0.5 if there is a minor gap; deduct 1 if the result is not obtained; do not go below 0.

2. $\mathrm{E}[R]$ $=$ ${\displaystyle \sum _{r=0}^n}r{p}_R(r)\mathit{\hspace{1em}}\text{def}\mathrm{E}$

   $=$ $0+{\displaystyle \sum _{r=1}^n}r{p}_R(r)\mathit{\hspace{1em}}\text{cunning}$

   $=$ ${\displaystyle \sum _{r=1}^n}r\left({\displaystyle \genfrac{}{}{0pt}{}{n}{r}}\right){\theta }^r{(1-\theta )}^{n-r}\mathit{\hspace{1em}}\text{subst.}$

   $=$ ${\displaystyle \sum _{r=1}^n}r{\displaystyle \frac{n!}{r!(n-r)!}}{\theta }^r{(1-\theta )}^{n-r}\mathit{\hspace{1em}}$

   $=$ ${\displaystyle \sum _{r=1}^n}{\displaystyle \frac{n!}{(r-1)!(n-r)!}}{\theta }^r{(1-\theta )}^{n-r}\mathit{\hspace{1em}}\text{cancel}$

   $=$ $n\theta {\displaystyle \sum _{r=1}^n}{\displaystyle \frac{(n-1)!}{(n-r)!(r-1)!}}{\theta }^{r-1}{(1-\theta )}^{n-r}\mathit{\hspace{1em}}$

   $=$ $n\theta {\displaystyle \sum _{s=0}^{n-1}}{\displaystyle \frac{(n-1)!}{s!(n-1-s)!}}{\theta }^s{(1-\theta )}^{n-1-s}\mathit{\hspace{1em}}\text{subst s=r-1}$

   $=$ $n\theta {\displaystyle \sum _{s=0}^{n-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{s}}\right){\theta }^s{(1-\theta )}^{n-1-s}$

   $=$ $n\theta {\displaystyle \sum _{s=0}^{n-1}}{p}_S(s)\mathit{\hspace{1em}}\text{where }S\sim \mathrm{Bin}(n-1,\theta )$

   $=$ $n\theta \times 1=n\theta ,$

   since summation of binomial pmf is 1.

   [2 marks]

   Similarly

   $\mathrm{E}[R(R-1)]$ $=$ ${\displaystyle \sum _{r=0}^n}r(r-1)p(r)\mathit{\hspace{1em}}\text{E of function}$

   $=$ $0+0+{\displaystyle \sum _{r=2}^n}r(r-1)p(r)\mathit{\hspace{1em}}\text{first two terms 0}$

   $=$ ${\displaystyle \sum _{r=2}^n}r(r-1){\displaystyle \frac{n!}{r!(n-r)!}}{\theta }^r{(1-\theta )}^{n-r}$

   $=$ ${\displaystyle \sum _{r=2}^n}{\displaystyle \frac{n!}{(r-2)!(n-r)!}}{\theta }^r{(1-\theta )}^{n-r}\mathit{\hspace{1em}}\text{cancel}$

   $=$ $n(n-1){\theta }^2{\displaystyle \sum _{r=2}^n}{\displaystyle \frac{(n-2)!}{(r-2)!((n-2)-(r-2))!}}{\theta }^{r-2}{(1-\theta )}^{(n-2)-(r-2)}\mathit{\hspace{1em}}\text{as before}$

   $=$ $n(n-1){\theta }^2{\displaystyle \sum _{s=0}^{n-2}}{\displaystyle \frac{(n-2)!}{s!((n-2)-s)!}}{\theta }^2{(1-\theta )}^{(n-2)-s}\mathit{\hspace{1em}}\text{subst s=r-2}$

   $=$ $n(n-1){\theta }^2{\displaystyle \sum _{s=0}^{n-2}}{p}_S(s)\mathit{\hspace{1em}}\text{where }S\sim \mathrm{Bin}(n-2,\theta )$

   $=$ $n(n-1){\theta }^2$

   since the binomial pmf sums to 1.

   [1]

   Hence

   $\mathrm{Var}(R)$ $=$ $\mathrm{E}[R(R-1)]+\mathrm{E}[R]-{(\mathrm{E}[R])}^2$

   $=$ $n(n-1){\theta }^2+n\theta -n^2{\theta }^2$

   $=$ $n\theta (1-\theta ).$

   [1]

3. The number abandoned is probably better modelled as a binomial rv with parameters the number of cars, and the probability that any individual car is abandone. However, we don’t know either of these quantities. But since we have a rare event, we can use Poisson with parameter $\lambda =2.2$ given by the average number.

   [1]

   Therefore

   1. $\mathrm{P}(R=0)=e^{-2.2}=0.1108$,

      [1]

   2. $\mathrm{P}(R\ge 2)=1-e^{-2.2}-(2.2)e^{-2.2}=0.6454$.

      [1]