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Refer to Exercise 5.88. If Y denotes the number of tosses
Chapter 5, Problem 115E(choose chapter or problem)
Refer to Exercise 5.88. If \(Y\) denotes the number of tosses of the die until you observe each of the six faces,\(Y=Y_{1}+Y_{2}+Y_{3}+Y_{4}+Y_{5}+Y_{6}\) where \(Y_{1}\) is the trial on which the first face is tossed, \(Y_{1}=1\), \(Y_{2}\) is the number of additional tosses required to get a face different than the first, \(Y_{3}\) is the number of additional tosses required to get a face different than the first two distinct faces, .... \(Y_{6}\) is the number of additional tosses to get the last remaining face after all other faces have been observed.
a Show that \(\operatorname{Cov}\left(Y_{i}, Y_{j}\right)=0, i, j=1,2, \ldots, 6, i \neq j\).
b Use Theorem 5.12 to find \(V(Y)\).
c Give an interval that will contain \(Y\) with probability at least 3/4.
Equation Transcription:
Text Transcription:
Y
Y=Y_1+Y_2+Y_3+Y_4+Y_5+Y_6
Y_1
Y_1=1
Y_2
Y_3
Y_6
Cov(Y_i,Y_j)=0,i,j=1,2,...6,i neq j
V(Y)
Y
Questions & Answers
QUESTION:
Refer to Exercise 5.88. If \(Y\) denotes the number of tosses of the die until you observe each of the six faces,\(Y=Y_{1}+Y_{2}+Y_{3}+Y_{4}+Y_{5}+Y_{6}\) where \(Y_{1}\) is the trial on which the first face is tossed, \(Y_{1}=1\), \(Y_{2}\) is the number of additional tosses required to get a face different than the first, \(Y_{3}\) is the number of additional tosses required to get a face different than the first two distinct faces, .... \(Y_{6}\) is the number of additional tosses to get the last remaining face after all other faces have been observed.
a Show that \(\operatorname{Cov}\left(Y_{i}, Y_{j}\right)=0, i, j=1,2, \ldots, 6, i \neq j\).
b Use Theorem 5.12 to find \(V(Y)\).
c Give an interval that will contain \(Y\) with probability at least 3/4.
Equation Transcription:
Text Transcription:
Y
Y=Y_1+Y_2+Y_3+Y_4+Y_5+Y_6
Y_1
Y_1=1
Y_2
Y_3
Y_6
Cov(Y_i,Y_j)=0,i,j=1,2,...6,i neq j
V(Y)
Y
ANSWER:
Solution :
Step 1 of 3:
Given we are told to toss a die until we have observed each of the 6 faces.
Let Y denotes the number of trials to complete the assignment.
So .
Where = the trial on which the first face is tossed and so on
= the number of additional tosses to get the last remaining face after all other faces have been observed.
Our goal is:
a). We need to find .
b). We need to find V(Y).
c). We need to find an interval that will contain Y with probability at least .
a).
Now we have to show that .
The covariance is zero, because every toss is independent of all other tosses and thus and
are independent if