Solution to Homework for Section 6.4, page 445
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5.  page 445

I did this one in class,  so refere to your class notes.
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7.  page  445

I did this one in class,  so refere to your class notes.
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11.  page 445

Refer to page 286 of the text for thr formulas that are used below.

T_{r}  =  W_{1}  +  ...  +  W_{r}

E( T_{r} )  =  r / lambda

sigma( T_{r} )  =  sqrt(r) / lambda

Thus,

E(T_{1}) = 1 / lambda,    sigma(T_{1}) = 1 / lambda

E(T_{3}) = 3 / lambda,    sigma(T_{1}) = sqrt(3) / lambda

We have,

Corr (T_1, T_3)  =  Cov(T_1, T_3) / sigma(T_1)sigma(T_3)

So,  what is left is to calculate  Cov(T_{1}, T_{3}).

Cov(T_{1}, T_{3})  =  E(T_{1} T_{3}) - E(T_{1}) E(T_{3})

                                  =  E(T_{1} T_{3}) - 3 / lambda^{2}

Hence,  we need to find E(T_{1} T_{3}).  But,

T_{1} T_{3}  =  W_{1} (W_{1} + W_{2} + W_{3})

                       =  W_{1}^{2} + W_{1} W_{2} + W_{1} W_{3}

Therefore,  using independence of the waiting times random variables

E(T_{1} T_{3}) = E(W_{1}^{2})  + 

                                E(W_{1})   E(W_{2})  + 

                                E(W_{1})   E(W_{3})

                             =  2/lambda^{2} + 1/lambda^{2} + 1/lambda^{2}

                             =  4/lambda^{2}

Note that

E(W_{1}^{2})=E(T_{1}^{2}=Var(T_1)+E(T_1)^{2}=1/lambda^{2}.

Finally,  we can put everything together

Cov(T_{1}, T_{3}) = 4/lambda^{2} - 3/lambda^{2} = 1/lambda^{2} 

=>  Corr(T_{1}, T_{3}) = (1/lambda^{2})  /  (sqrt(3)/lambda^{2})

                                         =  1 / sqrt(3).
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