4.5 Quiz 5

Figure 4.41: Problem 1

(a) Consider

\begin{align*} M_{W}\left ( t\right ) & =E\left ( e^{wt}\right ) \\ & =E\left ( e^{(u+v)t}\right ) =E\left ( e^{ut}e^{vt}\right ) \end{align*}

But since \(U\perp V\) then the above reduces to

\begin{align*} M_{W}\left ( t\right ) & =E\left ( e^{ut}\right ) E\left ( e^{vt}\right ) \\ & =M_{U}\left ( t\right ) M_{V}\left ( t\right ) \end{align*}

Hence

\begin{align*} M_{U}\left ( t\right ) & =\frac {M_{W}\left ( t\right ) }{M_{U}\left ( t\right ) }=\frac {\left ( 1-2t\right ) ^{-\frac {n}{2}}}{\left ( 1-2t\right ) ^{-\frac {1}{2}}}\\ & =\left ( 1-2t\right ) ^{-\frac {n}{2}+\frac {1}{2}}\\ & =\left ( 1-2t\right ) ^{-\frac {\left ( n-1\right ) }{2}}\end{align*}

Hence

\[ U \chi ^{2} \text {with} \left ( n-1\right ) \text {degrees of freedom} \]

(b)

\begin{align*} E\left ( W\right ) & =E\left ( U+V\right ) \\ & =E\left ( U\right ) +E\left ( V\right ) \end{align*}

Now use the moment generation function to find the expectations of \(U\) and \(V.\)

Need to find \(M_{X}^{\prime }\left ( t\right ) \) where \(M_{X}\left ( t\right ) =\left ( 1-2t\right ) ^{-\frac {m}{2}}\) where \(m\) is the degree of freedom

\begin{align*} \frac {d}{dt}M_{X}\left ( t\right ) & =\frac {d}{dt}\left ( 1-2t\right ) ^{-\frac {m}{2}}\\ & =-\frac {m}{2}\left ( 1-2t\right ) ^{-\frac {m}{2}-1}\left ( -2\right ) \end{align*}

Hence

\begin{equation} M^{\prime }\left ( t\right ) =m\left ( 1-2t\right ) ^{-\frac {m}{2}-1} \tag {1}\end{equation}

at \(t=0\) the above becomes

\[ M_{X}^{\prime }\left ( 0\right ) =m \]

For \(U\), we found that \(m=\left ( n-1\right ) \), hence

\[ \fbox {$E\left ( U\right ) =\left ( n-1\right ) $}\ \]

and for \(V\) we are told its degree of freedom is \(m=1\) hence

\[ E\left ( V\right ) =1 \]

Therefore

\[ E\left ( W\right ) =\left ( n-1\right ) +1 \]

Hence

\[ E\left ( W\right ) =n \]

Now

\begin{align*} Var(W) & =E\left ( W^{2}\right ) -\left [ E\left ( W\right ) \right ] ^{2}\\ & =E\left ( W^{2}\right ) -n^{2}\\ & =E\left ( \left ( U+V\right ) ^{2}\right ) -n^{2}\\ & =E\left ( \left ( U^{2}+V^{2}+2UV\right ) \right ) -n^{2}\\ & =E\left ( U^{2}\right ) +E\left ( V^{2}\right ) +2E\left ( UV\right ) -n^{2}\end{align*}

But \(U\perp V\) so the above becomes

\[ Var(W)=E\left ( U^{2}\right ) +E\left ( V^{2}\right ) +2E\left ( U\right ) E\left ( V\right ) -n^{2} \]

Lets find\(\ E\left ( Z^{2}\right ) \) for a \(Z\) chi square random variable of degree of freedom \(m\). We already found \(M^{\prime }\left ( t\right ) \) above in (1)

\begin{align*} E\left ( Z^{2}\right ) & =\left . M_{Z}^{^{\prime \prime }}\left ( t\right ) \right \vert _{t=0}\\ & =\frac {d}{dt}\left ( M_{Z}^{\prime }\left ( t\right ) \right ) \\ & =\frac {d}{dt}\left ( m\left ( 1-2t\right ) ^{-\frac {m}{2}-1}\right ) \\ & =m\left ( \left ( -\frac {m}{2}-1\right ) \left ( 1-2t\right ) ^{-\frac {m}{2}-2}\left ( -2\right ) \right ) \end{align*}

At \(t=0\)

\[ E\left ( Z^{2}\right ) =-2m\left ( -\frac {m}{2}-1\right ) \]

Hence

\begin{equation} E\left ( Z^{2}\right ) =m\left ( m+2\right ) \tag {2}\end{equation}

Hence using (2) above, we now can find \(E\left ( U^{2}\right ) \) and \(E\left ( V^{2}\right ) \)

For \(U\) it has degree of freedom \(m=\left ( n-1\right ) \), hence

\begin{align*} E(U^{2}) & =\left ( n-1\right ) \left ( \left ( n-1\right ) +2\right ) \\ & =n^{2}-1 \end{align*}

For \(V\) it has degree of freedom \(m=1\), hence

\begin{align*} E(V^{2}) & =1\times \left ( 1+2\right ) \\ & =3 \end{align*}

Hence

\begin{align*} Var(W) & =\left ( n^{2}-1\right ) +3+2\left ( n-1\right ) \times 1-n^{2}\\ & =n^{2}-1+3+2n-2-n^{2}\end{align*}

Hence

\[ Var(W)=2n \]

Figure 4.42: Problem 2

\[ X\left ( x\right ) \frac {\lambda ^{x}e^{-\lambda }}{x!} \]

Moment generating function for a Poisson r.v. or parameter \(\lambda \) is (from page 144)

\[ \fbox {$M_{X}\left ( t\right ) =e^{-\lambda }e^{e^{t}\lambda }$}\]

Now

\begin{align*} M_{Y}\left ( t\right ) & =E\left ( e^{yt}\right ) \\ & =E\left ( e^{\sqrt {x}t}\right ) \end{align*}

Hence

\[ M_{Y}^{\prime }\left ( t\right ) =E\left ( \sqrt {x}e^{\sqrt {x}t}\right ) \]

and

\begin{align*} M_{Y}^{\prime \prime }\left ( t\right ) & =E\left ( \sqrt {x}\sqrt {x}e^{\sqrt {x}t}\right ) \\ & =E\left ( xe^{\sqrt {x}t}\right ) \end{align*}

Therefore

\[ M_{Y}^{\prime \prime }\left ( 0\right ) =E\left ( x\right ) \]

But\(\ \)

\begin{align*} E\left ( x\right ) & =\left . M_{X}^{\prime }\left ( t\right ) \right \vert _{t=0}\\ & =\lambda \end{align*}

Hence

\[ M_{Y}^{\prime \prime }\left ( 0\right ) =\lambda \]

But

\[ M_{Y}^{\prime \prime }\left ( 0\right ) =E\left ( Y^{2}\right ) \]

then

\[ \fbox {$E\left ( Y^{2}\right ) =\lambda $}\]

Now to find \(Var\left ( Y\right ) \)

\[ Var\left ( Y\right ) =E\left ( Y^{2}\right ) -\left [ E\left ( Y\right ) \right ] ^{2}\]

Where

\begin{align*} E\left ( Y\right ) & =M_{Y}^{\prime }\left ( 0\right ) \\ & =E\left ( \sqrt {x}\right ) \end{align*}

So we need to find \(E\left ( \sqrt {x}\right ) \) to complete the solution.

\begin{align*} E\left ( \sqrt {x}\right ) & =\sum _{x=0}^{\infty }\sqrt {x}\frac {\lambda ^{x}e^{-\lambda }}{x!}\\ & =0+\frac {\lambda e^{-\lambda }}{1!}+\sqrt {2}\frac {\lambda ^{2}e^{-\lambda }}{2!}+\sqrt {3}\frac {\lambda ^{3}e^{-\lambda }}{3!}+\sqrt {4}\frac {\lambda ^{4}e^{-\lambda }}{4!}+\cdots \\ & =e^{-\lambda }\left ( \lambda +\sqrt {2}\frac {\lambda ^{2}}{2!}+\sqrt {3}\frac {\lambda ^{3}}{3!}+\sqrt {4}\frac {\lambda ^{4}}{4!}+\cdots \right ) \\ & =\lambda e^{-\lambda }\left ( 1+\frac {1}{\sqrt {2}}\lambda +\frac {1}{\sqrt {3}}\frac {\lambda ^{2}}{2!}+\frac {1}{\sqrt {4}}\frac {\lambda ^{3}}{3!}+\frac {1}{\sqrt {5}}\frac {\lambda ^{4}}{4!}+\cdots \right ) \\ & =\lambda e^{-\lambda }\left ( 1+\frac {\lambda }{\sqrt {2}}+\frac {2}{\sqrt {3}}\frac {\left ( \frac {\lambda }{\sqrt {2}}\right ) ^{2}}{2!}+\frac {2\sqrt {2}}{\sqrt {4}}\frac {\left ( \frac {\lambda }{\sqrt {2}}\right ) ^{3}}{3!}+\frac {4}{\sqrt {5}}\frac {\left ( \frac {\lambda }{\sqrt {2}}\right ) ^{4}}{4!}+\cdots \right ) \\ & =\lambda e^{-\lambda }\left ( 1+\frac {\lambda }{\sqrt {2}}+1.\,\allowbreak 154\,7\frac {\left ( \frac {\lambda }{\sqrt {2}}\right ) ^{2}}{2!}+1.\,\allowbreak 414\,2\frac {\left ( \frac {\lambda }{\sqrt {2}}\right ) ^{3}}{3!}+1.\,\allowbreak 788\,9\frac {\left ( \frac {\lambda }{\sqrt {2}}\right ) ^{4}}{4!}+\cdots \right ) \\ & \simeq \lambda e^{-\lambda }\left ( e^{\frac {\lambda }{\sqrt {2}}}\right ) \\ & =\lambda e^{\frac {\lambda }{\sqrt {2}}-\lambda }=\lambda e^{\frac {\lambda (1-\sqrt {2)}}{\sqrt {2}}}\end{align*}

Hence

\begin{align*} Var\left ( Y\right ) & =\lambda -\left [ \lambda e^{\frac {\lambda (1-\sqrt {2)}}{\sqrt {2}}}\right ] ^{2}\\ & =\lambda -\lambda ^{2}e^{\frac {2\lambda (1-\sqrt {2)}}{\sqrt {2}}}\\ & =\lambda -\lambda ^{2}e^{\sqrt {2}\lambda (1-\sqrt {2)}}\end{align*}

Hence

\[ Var\left ( Y\right ) \simeq \lambda \left ( 1-\lambda e^{-0.585\,78\lambda }\right ) \]

Figure 4.43: Problem 3

(a)

\begin{align*} f_{Y}\left ( y\right ) & =\frac {\lambda ^{\alpha }}{\Gamma \left ( \alpha \right ) }y^{\alpha -1}e^{-\lambda y}\ \ \ \ \ y\geq 0\\ f_{\left ( X|Y=y\right ) }\left ( x|y\right ) & =\frac {\left ( y^{2}\right ) ^{x}e^{-y^{2}}}{y!}\ \ \ \ \ \ \ x=0,1,2,\cdots \end{align*}

Now

\[ E\left ( X\right ) =E\left ( E\left ( X|Y\right ) \right ) \]

But \(E\left ( X|Y\right ) \) is expectation of a Poisson r.v. with parameter \(Y^{2}\). But we know that mean of a poisson r.v. with parameter \(\lambda \) is \(\lambda \). Hence \(E\left ( X|Y\right ) =Y^{2}\) since we are told \(Y^{2}\) is the parameter.

Hence

\[ E\left ( X\right ) =E\left ( Y^{2}\right ) \]

But the moment generating function for Gamma is \(M_{Y}\left ( t\right ) =\left ( \frac {\lambda }{\lambda -t}\right ) ^{\alpha }\) (book page 145 second edition).

Hence \(E\left ( Y^{2}\right ) =M_{Y}^{\prime \prime }\left ( 0\right ) =\frac {\alpha \left ( \alpha +1\right ) }{\lambda ^{2}}\) (page 145)

Hence

\[ \fbox {$E\left ( X\right ) =\frac {\alpha \left ( \alpha +1\right ) }{\lambda ^{2}}$}\]

(b)

\begin{equation} Var\left ( X\right ) =E\left ( X^{2}\right ) -\left [ E\left ( X\right ) \right ] ^{2}\tag {1}\end{equation}

But

\[ E\left ( X^{2}\right ) =E\left ( E\left ( X^{2}|Y\right ) \right ) \]

But \(E\left ( X^{2}|Y\right ) \) is \(E\left ( X^{2}\right ) \) of a poisson r.v. with parameter \(Y^{2}\). But we know that \(E\left ( X^{2}\right ) \) of a poisson r.v. with parameter \(\lambda \) is \(\lambda ^{2}+\lambda \) (book page 144 example A). Hence since we are told \(Y^{2}\) is the parameter, then

\begin{align*} E\left ( X^{2}|Y\right ) & =\left ( Y^{2}\right ) ^{2}+Y^{2}\\ & =\left ( Y^{4}+Y^{2}\right ) \end{align*}

Hence

\begin{align*} E\left ( X^{2}\right ) & =E\left ( Y^{4}+Y^{2}\right ) \\ & =E\left ( Y^{4}\right ) +E\left ( Y^{2}\right ) \end{align*}

But using mgf for Gamma distribution we can find \(E\left ( Y^{4}\right ) \).

\begin{align*} M_{Y}^{\prime \prime \prime \prime }\left ( t\right ) & =\frac {d^{4}}{dt^{4}}\left ( \frac {\lambda }{\lambda -t}\right ) ^{\alpha }\\ & =\frac {\alpha \left ( 6+11\alpha +6\alpha ^{2}+\alpha ^{3}\right ) }{\left ( t-\lambda \right ) ^{4}}\left ( \frac {\lambda }{\lambda -t}\right ) ^{\alpha }\end{align*}

Then

\[ M_{Y}^{\prime \prime \prime \prime }\left ( 0\right ) =\frac {\alpha \left ( 6+11\alpha +6\alpha ^{2}+\alpha ^{3}\right ) }{\lambda ^{4}}\]

Therefore

\[ E\left ( X^{2}\right ) =\frac {\alpha \left ( 6+11\alpha +6\alpha ^{2}+\alpha ^{3}\right ) }{\lambda ^{4}}+\frac {\alpha \left ( \alpha +1\right ) }{\lambda ^{2}}\]

Then (1) becomes

\begin{align*} Var\left ( X\right ) & =\left ( \frac {\alpha \left ( 6+11\alpha +6\alpha ^{2}+\alpha ^{3}\right ) }{\lambda ^{4}}+\frac {\alpha \left ( \alpha +1\right ) }{\lambda ^{2}}\right ) -\left ( \frac {\alpha \left ( \alpha +1\right ) }{\lambda ^{2}}\right ) ^{2}\\ & =\frac {\alpha \left ( 6+11\alpha +6\alpha ^{2}+\alpha ^{3}\right ) }{\lambda ^{4}}+\frac {\alpha \left ( \alpha +1\right ) }{\lambda ^{2}}-\frac {\alpha ^{2}\left ( \alpha +1\right ) ^{2}}{\lambda ^{4}}\end{align*}

Then

\[ \fbox {$Var\left ( X\right ) =\frac {\alpha }{\lambda ^{4}}\left ( \alpha +1\right ) \left ( \lambda ^{2}+4\alpha +6\right ) $}\]

4.5.1 Graded

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