4.2
By definition,
\[ \Gamma \left ( z\right ) =\int _{0}^{\infty }t^{z-1}e^{-t}dt \]
Hence
\[ \Gamma \left ( \frac {n}{2}\right ) =\int _{0}^{\infty }t^{\left ( \frac {n}{2}-1\right ) }e^{-t}dt \]
When \(n=1\) , we are told that
\[ \Gamma \left ( \frac {1}{2}\right ) =\int _{0}^{\infty }t^{\left ( \frac {1}{2}-1\right ) }e^{-t}dt=\int _{0}^{\infty }t^{\left ( -\frac {1}{2}\right ) }e^{-t}dt=\sqrt {\pi }\]
For \(n=3\,\) , we have
\begin{align*} \Gamma \left ( \frac {3}{2}\right ) & =\int _{0}^{\infty }t^{\left ( \frac {3}{2}-1\right ) }e^{-t}dt\\ & =\int _{0}^{\infty }t^{\left ( \frac {1}{2}\right ) }e^{-t}dt \end{align*}
Now do integration by parts,
\begin{align*} \Gamma \left ( \frac {3}{2}\right ) & =\int _{0}^{\infty }\overset {u}{\overbrace {t^{\left ( \frac {1}{2}\right ) }}}\overset {dv}{\overbrace {e^{-t}dt}}\\ & =\left [ t^{\frac {1}{2}}\left ( -e^{-t}\right ) \right ] _{0}^{\infty }-\int _{0}^{\infty }\frac {1}{2}t^{\left ( -\frac {1}{2}\right ) }\left ( -e^{-t}\right ) dt\\ & =-\left [ t^{\frac {1}{2}}e^{-t}\right ] _{0}^{\infty }+\int _{0}^{\infty }\frac {1}{2}t^{\left ( -\frac {1}{2}\right ) }e^{-t}dt \end{align*}
But \(\left [ t^{\frac {1}{2}}e^{-t}\right ] _{0}^{\infty }=\left [ 0-0\right ] =0\) and the above becomes
\[ \Gamma \left ( \frac {3}{2}\right ) =\frac {1}{2}\int _{0}^{\infty }t^{\left ( -\frac {1}{2}\right ) }e^{-t}dt \]
But \(\int _{0}^{\infty }t^{\left ( -\frac {1}{2}\right ) }e^{-t}dt=\Gamma \left ( \frac {1}{2}\right ) \) , hence the above becomes
\[ \fbox {$\Gamma \left ( \frac {3}{2}\right ) =\frac {1}{2}\Gamma \left ( \frac {1}{2}\right ) $}\]
Now do the same for \(n=5\)
\begin{align*} \Gamma \left ( \frac {5}{2}\right ) & =\int _{0}^{\infty }t^{\left ( \frac {5}{2}-1\right ) }e^{-t}dt\\ & =\int _{0}^{\infty }\overset {u}{\overbrace {t^{\left ( \frac {3}{2}\right ) }}}\overset {dv}{\overbrace {e^{-t}dt}}\\ & =-\left [ t^{\frac {3}{2}}e^{-t}\right ] _{0}^{\infty }-\int _{0}^{\infty }\frac {3}{2}t^{\left ( \frac {1}{2}\right ) }\left ( -e^{-t}\right ) dt\\ & =\frac {3}{2}\int _{0}^{\infty }t^{\left ( \frac {1}{2}\right ) }e^{-t}dt \end{align*}
But \(\int _{0}^{\infty }t^{\left ( \frac {1}{2}\right ) }e^{-t}dt\) we found from above to be \(\Gamma \left ( \frac {3}{2}\right ) \) , hence
\[ \Gamma \left ( \frac {5}{2}\right ) =\frac {3}{2}\Gamma \left ( \frac {3}{2}\right ) \]
But \(\Gamma \left ( \frac {3}{2}\right ) =\frac {1}{2}\Gamma \left ( \frac {1}{2}\right ) \) from above, hence
\begin{align*} \Gamma \left ( \frac {5}{2}\right ) & =\frac {3}{2}\frac {1}{2}\Gamma \left ( \frac {1}{2}\right ) \\ & =\frac {3}{2}\frac {1}{2}\sqrt {\pi }\end{align*}
Continuing this way, we find that \(\Gamma \left ( \frac {7}{2}\right ) =\frac {5}{2}\frac {3}{2}\frac {1}{2}\Gamma \left ( \frac {1}{2}\right ) \) , and hence in general
\begin{equation} \Gamma \left ( \frac {n}{2}\right ) =\frac {\left ( n-2\right ) }{2}\frac {\left ( n-4\right ) }{2}\frac {\left ( n-6\right ) }{2}\cdots \frac {5}{2}\frac {3}{2}\frac {1}{2}\sqrt {\pi } \tag {1}\end{equation}
Now,
\[ \left ( n-1\right ) !=\left ( n-1\right ) \left ( n-2\right ) \left ( n-3\right ) \left ( n-4\right ) \left ( n-5\right ) \left ( n-6\right ) \cdots 5\times 4\times 3\times 2\times 1 \]
Hence from above we see that
\[ \left ( n-2\right ) \left ( n-4\right ) \left ( n-6\right ) \cdots 5\times 3\times 1=\frac {\left ( n-1\right ) !}{\left ( n-1\right ) \left ( n-3\right ) \left ( n-5\right ) \cdots 4\times 2\times 1}\]
Therefore (1) can be written as
\begin{align} \Gamma \left ( \frac {n}{2}\right ) & =\left ( \frac {\left ( n-1\right ) !}{\left ( n-1\right ) \left ( n-3\right ) \left ( n-5\right ) \cdots 4\times 2\times 1}\right ) \overset {\text {There are }\frac {n-1}{2}\text { such terms}}{\overbrace {\left ( \frac {1}{2}\frac {1}{2}\frac {1}{2}\cdots \frac {1}{2}\frac {1}{2}\right ) }}\sqrt {\pi }\nonumber \\ & =\frac {\left ( n-1\right ) !}{\left ( n-1\right ) \left ( n-3\right ) \left ( n-5\right ) \cdots 4\times 2}\frac {1}{2^{\left ( \frac {n-1}{2}\right ) }}\sqrt {\pi } \tag {2}\end{align}
But
\[ 2^{\left ( \frac {n-1}{2}\right ) }=\left ( 2^{n-1}\right ) \left ( 2^{-\frac {n-1}{2}}\right ) =\frac {2^{n-1}}{2^{\frac {n-1}{2}}}\]
Hence (2) becomes
\[ \Gamma \left ( \frac {n}{2}\right ) =\left ( \frac {\left ( n-1\right ) !}{\frac {1}{2^{\frac {n-1}{2}}}\left ( n-1\right ) \left ( n-3\right ) \left ( n-5\right ) \cdots 4\times 2}\right ) \frac {1}{2^{n-1}}\sqrt {\pi }\]
But there are \(\frac {n-1}{2}\) terms in the expression \(\left ( n-1\right ) \left ( n-3\right ) \left ( n-5\right ) \cdots 4\times 2\) in the denominator above and we have \(\frac {n-1}{2}\) number of \(\frac {1}{2}\) sitting
there, which we can distribute now below each terms to obtain
\begin{equation} \Gamma \left ( \frac {n}{2}\right ) =\left ( \frac {\left ( n-1\right ) !}{\frac {\left ( n-1\right ) }{2}\frac {\left ( n-3\right ) }{2}\frac {\left ( n-5\right ) }{2}\cdots \frac {4}{2}\times \frac {2}{2}}\right ) \frac {1}{2^{n-1}}\sqrt {\pi } \tag {3}\end{equation}
But
\begin{align*} \left ( \frac {n-1}{2}\right ) ! & =\left ( \frac {n-1}{2}\right ) \left ( \frac {n-1}{2}-1\right ) \left ( \frac {n-1}{2}-2\right ) \cdots \times 4\times 3\times 2\times 1\\ & =\left ( \frac {n-1}{2}\right ) \left ( \frac {n-3}{2}\right ) \left ( \frac {n-5}{2}\right ) \cdots \times 4\times 2 \end{align*}
Compare the above to the denominator term in (3) we see it is the same. Hence (3) can be written
as
\[ \fbox {$\Gamma \left ( \frac {n}{2}\right ) =\frac {\left ( n-1\right ) !}{\left ( \frac {n-1}{2}\right ) !}\frac {1}{2^{n-1}}\sqrt {\pi }$}\]
Which is what we are asked to show.
Since \(U\) is continuous r.v., we start with the CDF of \(Z\)
\begin{align} F_{Z}\left ( z\right ) & =P\left ( Z\leq z\right ) \nonumber \\ & =P\left ( U^{2}\leq z\right ) \nonumber \\ & =P\left ( -\sqrt {z}\leq U\leq \sqrt {z}\right ) \tag {1}\end{align}
But since \(F_{U}^{\prime }\left ( u\right ) =f_{U}\left ( u\right ) \) , then we know that \(P\left ( a\leq U\leq b\right ) =\int _{a}^{b}f_{U}\left ( x\right ) dx\rightarrow F_{U}\left ( b\right ) -F_{U}\left ( a\right ) \)
Hence RHS of (1) becomes
\[ F_{Z}\left ( z\right ) =F_{U}\left ( \sqrt {z}\right ) -F_{U}\left ( -\sqrt {z}\right ) \]
Therefore, taking derivatives with respect to \(z\) we obtain
\begin{align*} f_{Z}\left ( z\right ) & =f_{U}\left ( \sqrt {z}\right ) \frac {d}{dz}\sqrt {z}-f_{U}\left ( -\sqrt {z}\right ) \frac {d}{dz}\left ( -\sqrt {z}\right ) \\ & =\frac {1}{2}z^{\frac {-1}{2}}f_{U}\left ( \sqrt {z}\right ) +f_{U}\left ( -\sqrt {z}\right ) \frac {1}{2}z^{\frac {-1}{2}}\\ & =\frac {1}{2}z^{\frac {-1}{2}}\left ( f_{U}\left ( \sqrt {z}\right ) +f_{U}\left ( -\sqrt {z}\right ) \right ) \end{align*}
Since \(U\) is uniform, hence \(f_{U}\left ( a\right ) =f_{U}\left ( -a\right ) \) , hence the above becomes
\begin{align*} f_{Z}\left ( z\right ) & =\frac {1}{2}z^{\frac {-1}{2}}\left ( 2f_{U}\left ( \sqrt {z}\right ) \right ) \\ & =\frac {1}{\sqrt {z}}\ f_{U}\left ( \sqrt {z}\right ) \end{align*}
Now I need to determine the limits of \(f_{Z}\left ( z\right ) \) and the shape. \(f_{U}\) is defined for real arguments from \(-1\) to \(+1\) . i.e. \(f_{U}\)
is real valued function of real arguments. Hence if \(z\) was negative then \(\sqrt {z}\) will be complex,
and so this will not be allowed. Hence we have to restrict \(z\geq 0\) . But now we observe that \(z=0\)
is not possible, since we will have \(\frac {1}{0}\) term, so this means \(z\) is strictly larger than zero.
So
\[ f_{Z}\left ( z\right ) =\frac {1}{\sqrt {z}}\ f_{U}\left ( \sqrt {z}\right ) \ \ \ \ \ z>0 \]
But we know that \(f_{U}\left ( x\right ) =\frac {1}{2}\) for up to \(x=1\) , hence this means when \(\sqrt {z}>1\) then \(f_{U}\left ( \sqrt {z}\right ) =0\) , when means when \(z>1\) then
\(f_{U}\left ( \sqrt {z}\right ) =0\)
Hence we now write
\[ \fbox {$f_{Z}\left ( z\right ) =\left \{ \begin {array} [c]{cc}\frac {1}{\sqrt {z}}\frac {1}{2} & \ 0<z\leq 1\\ 0 & z>1\\ undefined & z\leq 0 \end {array} \right . $}\]
Here is a plot
I explain the idea behind obtaining a discrete random number from a continues random number by
the following diagram below. We assume that the discrete random number belongs to some
distribution. In this example, we are told what the distribution is. We know that the CDF for
geometric random variable is given by
\[ F_{K}\left ( k\right ) =1-\left ( 1-p\right ) ^{k}\]
We see from the above diagram, that once we are given \(u\) we need to find \(k\) which satisfy the
following identity
\[ F_{K}\left ( k-1\right ) <u\leq F_{K}\left ( k\right ) \]
Or in other words
\[ 1-\left ( 1-p\right ) ^{k-1}<u\leq 1-\left ( 1-p\right ) ^{k}\]
The specific discrete value \(k\) which will satisfy the above, is the random variable we want, which
belong to the geometeric distribution.
Now when \(u=0.0153,\) and since \(p=\frac {1}{3}\) , we have
\[ 1-\left ( 1-\frac {1}{3}\right ) ^{k-1}<0.0153\leq 1-\left ( 1-\frac {1}{3}\right ) ^{k}\]
for \(k=1,\) we have
\[ 0<0.0153\leq 0.333\,33\ \ \ \ \ \ \ \ \ \ \ \ YES \]
Hence \(k=1\) is the random variable associated with \(u=0.0153\)
Now let us do \(u=0.7468\)
for \(k=1,\) we have
\[ 0<0.7468\leq 0.333\,33\ \ \ \ \ \ \ \ \ NO \]
try \(k=2\)
\begin{align*} 1-\left ( 1-\frac {1}{3}\right ) ^{1} & <0.7468\leq 1-\left ( 1-\frac {1}{3}\right ) ^{2}\\ 0.333\,33\ \ & <0.7468\leq 0.555\,56\ \ \ \ \ \ NO \end{align*}
try \(k=3\)
\begin{align*} 1-\left ( 1-\frac {1}{3}\right ) ^{2} & <0.7468\leq 1-\left ( 1-\frac {1}{3}\right ) ^{3}\\ 0.555\,56 & <0.7468\leq 0.703\,7\ \ \ \ \ \ \ NO \end{align*}
try \(k=4\)
\begin{align*} 1-\left ( 1-\frac {1}{3}\right ) ^{3} & <0.7468\leq 1-\left ( 1-\frac {1}{3}\right ) ^{4}\\ 0.703\,7\ & <0.7468\leq 0.802\,47\ \ \ \ \ \ \ YES \end{align*}
Hence \(k=4\) is the random variable associated with \(u=0.7468\)
Now let us do \(u=0.4451\)
We see from the above, that this will have \(k=2\) since for \(k=2\) the intervals is \(0.333\,33\ \ <u\leq 0.555\,56\ \)
Hence \(k=2\) is the random variable associated with \(u=0.4451\)
Now let us do \(u=0.9318\)
From above, we see that this will have a \(k\) larger than 4, so we do not need to try from the start, we
can start trying from \(k=5\)
try \(k=5\)
\begin{align*} 1-\left ( 1-\frac {1}{3}\right ) ^{4} & <0.9318\leq 1-\left ( 1-\frac {1}{3}\right ) ^{5}\\ 0.802\,47 & <0.9318\leq 0.868\,31\ \ \ \ \ \ \ NO \end{align*}
try \(k=6\)
\begin{align*} 1-\left ( 1-\frac {1}{3}\right ) ^{5} & <0.9318\leq 1-\left ( 1-\frac {1}{3}\right ) ^{6}\\ 0.868\,31 & <0.9318\leq 0.912\,21\ \ \ \ \ \ NO \end{align*}
try \(k=7\)
\begin{align*} 1-\left ( 1-\frac {1}{3}\right ) ^{6} & <0.9318\leq 1-\left ( 1-\frac {1}{3}\right ) ^{7}\\ 0.912\,21\ & <0.9318\leq 0.941\,47\ \ \ \ \ \ YES \end{align*}
Hence \(k=7\) is the random variable associated with \(u=0.9318\)
Hence result is
\(u\)
\(k\) \(0.0153\)
\(1\) \(0.4451\)
\(2\) \(0.7468\) \(4\)
\(0.9318\) \(7\)
of course one would write a program to do this.
(a)
Let \(P\left ( x_{i}=n_{i}\right ) \) means probability of player \(i\) winning \(n_{i}\) rounds.
We have 3 players, and a total of 10 rounds. Let the players be called \(x_{1},x_{2},x_{3}\) . Let the number of games
WON by \(x_{1}\) be \(n_{1}\) , and number of games won by \(x_{2}\) be \(n_{2}\) , and number of games won by \(x_{3}\) be
\(n_{3}\) .
Since we have 10 rounds, then we must have 10 wins as well. (some one must win). Hence we have
10 wins and 3 ways to split it, where each ’bucket’ is of different size. So this is a multi set
selection. called multinomial in the book using proposition B in chapter 1, we see that the total
number of ways the games can be won is \(\begin {pmatrix} 10\\ n_{1}n_{2}n_{3}\end {pmatrix} \)
But we need to find the probability of each one such combination. So we need to multiply the
above by the probability each player wins the number of the games they happened to win, which is
\(P\left ( x_{i}=n_{i}\right ) =p^{n_{i}}\) , but \(p=\frac {1}{3}\) for each player to win a round. Hence we write
\begin{align*} P\left ( x_{1}=n_{1},x_{2}=n_{2},x_{3}=n_{3}\right ) & =\begin {pmatrix} 10\\ n_{1}n_{2}n_{3}\end {pmatrix} \left ( \frac {1}{3}\right ) ^{n_{1}}\left ( \frac {1}{3}\right ) ^{n_{2}}\left ( \frac {1}{3}\right ) ^{n_{3}}\\ & =\frac {10!}{n_{1}!\ n_{2}!\ n_{3}!}\left ( \frac {1}{3}\right ) ^{n_{1}+n_{2}+n_{3}}\\ & =\frac {10!}{n_{1}!\ n_{2}!\ n_{3}!}\left ( \frac {1}{3}\right ) ^{10}\end{align*}
So the above is the joint probability that \(p_{1}\) wins \(n_{1}\) rounds and \(p_{2}\) wins \(n_{2}\) rounds and \(p_{3}\) wins \(n_{3}\)
rounds.
(b)We need to find \(P\left ( x_{1}=n_{1}\right ) ,\) i.e. the probability of first player winning \(n_{1}\) rounds.
\[ P\left ( x_{1}=n_{1}\right ) =\sum _{\substack {n_{2}=0,1,\cdots 10\\n_{3}=0,1,\cdots 10}}P\left ( x_{1}=n_{1},x_{2}=n_{2},x_{3}=n_{3}\right ) \]
To simplify, let me write \(P\left ( n_{1},n_{2},n_{3}\right ) \) instead, where the position of the \(n\) implies the player. So
\(p\left ( 0,1,9\right ) \) means player one wins zero rounds and player 2 wins 1 round and player 3 wins 9
rounds.
So the above becomes
\[ P\left ( x_{1}=n_{1}\right ) =\sum _{\substack {n_{2}=0,1,\cdots 10\\n_{3}=0,1,\cdots 10}}P\left ( n_{1},n_{2},n_{3}\right ) \]
But since \(n_{1}=10-\left ( n_{2}+n_{3}\right ) \) we see that we only need to count those terms in the above sum when this is true. i.e.
we do not need to count a term such as \(p\left ( 1,0,0\right ) \) since this is zero probability of happening. Now we
write
\[ P\left ( x_{1}=n_{1}\right ) =P\left ( n_{1},10-n_{1},0\right ) +P\left ( n_{1},9-n_{1},1\right ) +P\left ( n_{1},8-n_{1},2\right ) +\cdots P\left ( n_{1},0,10-n_{1}\right ) \]
For example,
\(P\left ( x_{1}=0\right ) =P\left ( 0,10,0\right ) +P\left ( 0,9,1\right ) +P\left ( 0,8,2\right ) +P\left ( 0,7,3\right ) +P\left ( 0,6,4\right ) +P\left ( 0,5,5\right ) +\)
\(P\left ( 0,4,6\right ) +P\left ( 0,3,7\right ) +P\left ( 0,2,8\right ) +P\left ( 0,1,9\right ) +P\left ( 0,0,10\right ) \)
But \(P\left ( 0,10,0\right ) =P\left ( 0,0,10\right ) \) and \(P\left ( 0,9,1\right ) =P\left ( 0,1,9\right ) \) , etc.. so the above can be written as
\begin{align*} P\left ( x_{1}=0\right ) & =2P\left ( 0,10,0\right ) +2P\left ( 0,9,1\right ) +2P\left ( 0,8,2\right ) +2P\left ( 0,7,3\right ) +2P\left ( 0,6,4\right ) +P\left ( 0,5,5\right ) \\ & \\ & =2\frac {10!}{0!\ 10!\ 0!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{0!\ 9!\ 1!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{0!\ 8!\ 2!}\left ( \frac {1}{3}\right ) ^{10}+\\ & 2\frac {10!}{0!7!\ 3!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{0!\ 6!\ 4!}\left ( \frac {1}{3}\right ) ^{10}+\frac {10!}{0!\ 5!\ 5!}\left ( \frac {1}{3}\right ) ^{10}\\ & \\ & =\fbox {$1.\,\allowbreak 734\,2\times 10^{-2}$}\end{align*}
and
\begin{align*} P\left ( x_{1}=1\right ) & =P\left ( 1,9,0\right ) +P\left ( 1,8,1\right ) +P\left ( 1,7,2\right ) +P\left ( 1,6,3\right ) +P\left ( 1,5,4\right ) +P\left ( 1,4,5\right ) \\ & +P\left ( 1,3,6\right ) +P\left ( 1,2,7\right ) +P\left ( 1,1,8\right ) +P\left ( 1,0,9\right ) \\ & \\ & =2P\left ( 1,9,0\right ) +2P\left ( 1,8,1\right ) +2P\left ( 1,7,2\right ) +2P\left ( 1,6,3\right ) +2P\left ( 1,5,4\right ) \\ & =2\frac {10!}{1!\ 9!\ 0!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{1!\ 8!\ 1!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{1!\ 7!\ 2!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{1!6!\ 3!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{1!\ 5!\ 4!}\left ( \frac {1}{3}\right ) ^{10}\\ & =\fbox {$8.\,\allowbreak 670\,8\times 10^{-2}$}\end{align*}
and
\begin{align*} P\left ( x_{1}=2\right ) & =P\left ( 2,8,0\right ) +P\left ( 2,7,1\right ) +P\left ( 2,6,2\right ) +P\left ( 2,5,3\right ) +P\left ( 2,4,4\right ) +P\left ( 2,3,5\right ) +\\ & P\left ( 2,2,6\right ) +P\left ( 2,1,7\right ) +P\left ( 2,0,8\right ) \\ & \\ & =2P\left ( 2,8,0\right ) +2P\left ( 2,7,1\right ) +2P\left ( 2,6,2\right ) +2P\left ( 2,5,3\right ) +P\left ( 2,4,4\right ) \\ & =2\frac {10!}{2!\ 8!\ 0!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{2!\ 7!\ 1!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{2!\ 6!\ 2!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{2!5!\ 3!}\left ( \frac {1}{3}\right ) ^{10}+\frac {10!}{2!\ 4!\ 4!}\left ( \frac {1}{3}\right ) ^{10}\\ & =\fbox {$0.195\,09$}\end{align*}
and
\begin{align*} P\left ( x_{1}=3\right ) & =P\left ( 3,7,0\right ) +P\left ( 3,6,1\right ) +P\left ( 3,5,2\right ) +P\left ( 3,4,3\right ) +P\left ( 3,3,4\right ) +\\ & P\left ( 3,2,5\right ) +P\left ( 3,1,6\right ) +P\left ( 3,0,7\right ) \\ & \\ & =2P\left ( 3,7,0\right ) +2P\left ( 3,6,1\right ) +2P\left ( 3,5,2\right ) +2P\left ( 3,4,3\right ) \\ & \\ & =2\frac {10!}{3!\ 7!\ 0!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{3!\ 6!\ 1!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{3!\ 5!\ 2!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{3!4!\ 3!}\left ( \frac {1}{3}\right ) ^{10}\\ & =\fbox {$0.260\,12$}\end{align*}
and
\begin{align*} P\left ( x_{1}=4\right ) & =P\left ( 4,6,0\right ) +P\left ( 4,5,1\right ) +P\left ( 4,4,2\right ) +P\left ( 4,3,3\right ) +P\left ( 4,2,4\right ) +P\left ( 4,1,5\right ) +P\left ( 4,0,6\right ) \\ & =2P\left ( 4,6,0\right ) +2P\left ( 4,5,1\right ) +2P\left ( 4,4,2\right ) +P\left ( 4,3,3\right ) \\ & \\ & =2\frac {10!}{4!\ 6!\ 0!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{4!\ 5!\ 1!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{4!\ 4!\ 2!}\left ( \frac {1}{3}\right ) ^{10}+\frac {10!}{4!3!\ 3!}\left ( \frac {1}{3}\right ) ^{10}\\ & =\fbox {$0.227\,61$}\end{align*}
and
\begin{align*} P\left ( x_{1}=5\right ) & =P\left ( 5,5,0\right ) +P\left ( 5,4,1\right ) +P\left ( 5,3,2\right ) +P\left ( 5,2,3\right ) +P\left ( 5,1,4\right ) +P\left ( 5,0,5\right ) \\ & =2P\left ( 5,5,0\right ) +2P\left ( 5,4,1\right ) +2P\left ( 5,3,2\right ) \\ & =2\frac {10!}{5!\ 5!\ 0!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{5!4!\ 1!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{5!\ 3!\ 2!}\left ( \frac {1}{3}\right ) ^{10}\\ & =\fbox {$0.136\,56$}\end{align*}
and
\begin{align*} P\left ( x_{1}=6\right ) & =P\left ( 6,4,0\right ) +P\left ( 6,3,1\right ) +P\left ( 6,2,2\right ) +P\left ( 6,1,3\right ) +P\left ( 6,0,4\right ) \\ & =2P\left ( 6,4,0\right ) +2P\left ( 6,3,1\right ) +P\left ( 6,2,2\right ) \\ & =2\frac {10!}{6!\ 4!\ 0!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{6!3!\ 1!}\left ( \frac {1}{3}\right ) ^{10}+\frac {10!}{6!\ 2!\ 2!}\left ( \frac {1}{3}\right ) ^{10}\\ & =\fbox {$5.\,\allowbreak 690\,2\times 10^{-2}$}\end{align*}
and
\begin{align*} P\left ( x_{1}=7\right ) & =P\left ( 7,3,0\right ) +P\left ( 7,2,1\right ) +P\left ( 7,1,2\right ) +P\left ( 7,0,3\right ) \\ & =2P\left ( 7,3,0\right ) +2P\left ( 7,2,1\right ) \\ & =2\frac {10!}{7!\ 3!\ 0!}\left ( \frac {1}{3}\right ) ^{10}+2\frac {10!}{7!2!\ 1!}\left ( \frac {1}{3}\right ) ^{10}\\ & =\fbox {$1.\,\allowbreak 625\,8\times 10^{-2}$}\end{align*}
and
\begin{align*} P\left ( x_{1}=8\right ) & =P\left ( 8,2,0\right ) +P\left ( 8,1,1\right ) +P\left ( 8,0,2\right ) \\ & =2P\left ( 8,2,0\right ) +P\left ( 8,1,1\right ) \\ & =2\frac {10!}{8!\ 2!\ 0!}\left ( \frac {1}{3}\right ) ^{10}+\frac {10!}{8!1!\ 1!}\left ( \frac {1}{3}\right ) ^{10}\\ & =\fbox {$3.\,\allowbreak 048\,3\times 10^{-3}$}\end{align*}
and
\begin{align*} P\left ( x_{1}=9\right ) & =P\left ( 9,1,0\right ) +P\left ( 9,0,1\right ) \\ & =2P\left ( 9,1,0\right ) \\ & =2\frac {10!}{9!\ 1!\ 0!}\left ( \frac {1}{3}\right ) ^{10}\\ & =\fbox {$3.\,\allowbreak 387\times 10^{-4}$}\end{align*}
and
\begin{align*} P\left ( x_{1}=10\right ) & =P\left ( 10,0,0\right ) \\ & =\frac {10!}{10!\ 0!\ 0!}\left ( \frac {1}{3}\right ) ^{10}\\ & =\fbox {$1.\,\allowbreak 693\,5\times 10^{-5}$}\end{align*}
Here is a plot of the marginal probability for player 1 winning \(n\) rounds
(a)
\begin{align*} f_{Y}\left ( y\right ) & =\int _{0}^{\infty }f\left ( x,y\right ) dx\\ & =\int _{0}^{\infty }\frac {1}{8}\left ( x^{2}-y^{2}\right ) e^{-x}dx \end{align*}
Integrate by parts, \(dv=e^{-x}dx,u=\left ( x^{2}-y^{2}\right ) \) , hence \(du=2x\) and \(v=-e^{-x}\) , so we obtain
\begin{align*} f_{Y}\left ( y\right ) & =\frac {1}{8}\left \{ \left [ \left ( x^{2}-y^{2}\right ) \left ( -e^{-x}\right ) \right ] _{0}^{\infty }-\int _{0}^{\infty }\left ( 2x\right ) \left ( -e^{-x}\right ) dx\right \} \\ & =\frac {1}{8}\left \{ -\left [ \left ( x^{2}-y^{2}\right ) \left ( e^{-x}\right ) \right ] _{0}^{\infty }+2\int _{0}^{\infty }xe^{-x}dx\right \} \\ & =\frac {1}{8}\left \{ -\left [ 0+y^{2}\right ] +2\int _{0}^{\infty }xe^{-x}dx\right \} \end{align*}
Do integration by parts again, \(dv=e^{-x}dx\) , \(u=x\) , hence
\begin{align*} f_{Y}\left ( y\right ) & =\frac {1}{8}\left \{ -y^{2}+2\left [ \left ( x\left ( -e^{-x}\right ) \right ) _{0}^{\infty }-\int _{0}^{\infty }-e^{-x}dx\right ] \right \} \\ & =\frac {1}{8}\left \{ -y^{2}+2\left [ -\left ( xe^{-x}\right ) _{0}^{\infty }+\int _{0}^{\infty }e^{-x}dx\right ] \right \} \\ & =\frac {1}{8}\left \{ -y^{2}+2\left [ 0+\left [ -e^{-x}\right ] _{0}^{\infty }\right ] \right \} \\ & =\frac {1}{8}\left \{ -y^{2}+2\left [ -\left [ 0-1\right ] \right ] \right \} \\ & =\frac {1}{8}\left \{ -y^{2}+2\right \} \end{align*}
Hence
\[ \fbox {$f_{Y}\left ( y\right ) =\frac {1}{8}\left ( 2-y^{2}\right ) $}\]
(b)The hard part is to determine the region to integrate. The following is the needed region which
satisfy \(P\left ( X+Y\leq 1\right ) \) and \(0\leq x\leq \infty \) and \(-x\leq y\leq x\)
For the top region,
\[ I_{1}=\int _{x=0}^{x=1}\int _{y=1-x}^{y=1}\frac {1}{8}\left ( x^{2}-y^{2}\right ) e^{-x}dydx \]
and for the bottom region
\[ I_{2}=\int _{x=1}^{x=2}\int _{y=0}^{y=-x}\frac {1}{8}\left ( x^{2}-y^{2}\right ) e^{-x}dydx \]
Hence
\[ \fbox {$P\left ( X+Y\leq 1\right ) =\int _{x=0}^{x=1}\int _{y=1-x}^{y=1}\frac {1}{8}\left ( x^{2}-y^{2}\right ) e^{-x}dydx+\int _{x=1}^{x=2}\int _{y=0}^{y=-x}\frac {1}{8}\left ( x^{2}-y^{2}\right ) e^{-x}dydx$}\]
4.2.1 14/20
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