Solution Found!
Suppose that Y is a binomial random variable based on n
Chapter 3, Problem 54E(choose chapter or problem)
Suppose that is a binomial random variable based on trials with success probability
and consider \(Y^{\star}=n-Y\).
a Argue that for \(y^{\star}=0,1, \ldots, n\)
\(P\left(Y^{\star}=y^{\star}\right)=P\left(n-Y=y^{\star}\right)=P\left(Y=n-y^{\star}\right)\).
b Use the result from part (a) to show that
\(P\left(Y^{\star}=y^{\star}\right)=\left(\begin{array}{c}
n \\
n-y^{\star}
\end{array}\right) p^{n-y^{\star}} q^{y^{\star}}=\left(\begin{array}{l}
n \\
y^{\star}
\end{array}\right) q^{y^{\star}} p^{n-y^{\star}}
\)
c The result in part (b) implies that \(Y^{\star}\) has a binomial distribution based on trials and "success" probability \(p^{\star}=q=1-p\). Why is this result "obvious"?
Equation Transcription:
Text Transcription:
Y=n-Y
y^star=0,1,...,n
P(Y^star=y^star)=P(n-Y=y^star)=P(Y=n-y^star)
P(Y^star=y^star)=(_n-y^star ^n)p^n-y^ starq^y^star=(_y^star ^n)q^y^starp^n-y^star
Y^star
p^star=q=1-p
Questions & Answers
QUESTION:
Suppose that is a binomial random variable based on trials with success probability
and consider \(Y^{\star}=n-Y\).
a Argue that for \(y^{\star}=0,1, \ldots, n\)
\(P\left(Y^{\star}=y^{\star}\right)=P\left(n-Y=y^{\star}\right)=P\left(Y=n-y^{\star}\right)\).
b Use the result from part (a) to show that
\(P\left(Y^{\star}=y^{\star}\right)=\left(\begin{array}{c}
n \\
n-y^{\star}
\end{array}\right) p^{n-y^{\star}} q^{y^{\star}}=\left(\begin{array}{l}
n \\
y^{\star}
\end{array}\right) q^{y^{\star}} p^{n-y^{\star}}
\)
c The result in part (b) implies that \(Y^{\star}\) has a binomial distribution based on trials and "success" probability \(p^{\star}=q=1-p\). Why is this result "obvious"?
Equation Transcription:
Text Transcription:
Y=n-Y
y^star=0,1,...,n
P(Y^star=y^star)=P(n-Y=y^star)=P(Y=n-y^star)
P(Y^star=y^star)=(_n-y^star ^n)p^n-y^ starq^y^star=(_y^star ^n)q^y^starp^n-y^star
Y^star
p^star=q=1-p
ANSWER:
Solution:
Step 1 of 3:
Let Y is a Binomial random variable based on n trials with success probability p.
Consider Y* = n - Y.
- The claim is to prove P(Y* = y*) = P(n - Y = y*) = P(Y = n - y*)
For y = 0, 1, 2,.......,n
We have Y* = n - Y
P(Y* = y*) = P(n - Y = y*)
Add Y on both sides
= P(n - Y + Y = y* + Y)
= P(n = y* + Y)
Subtract y* on both sides
P(Y* = y*) = P(n - y* = y* + Y - y*)
= P(n - y* = Y)
Interchange the left and right side
P(Y* = y*) = P(Y = n - y* )
Hence, P(Y* = y*) = P(n - Y = y*) = P(Y = n - y*)