Suppose that Y is a binomial random variable based on n

Chapter 3, Problem 54E

(choose chapter or problem)

Get Unlimited 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

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.

  1. 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*)


Add to cart


Study Tools You Might Need

Not The Solution You Need? Search for Your Answer Here:

×

Login

Login or Sign up for access to all of our study tools and educational content!

Forgot password?
Register Now

×

Register

Sign up for access to all content on our site!

Or login if you already have an account

×

Reset password

If you have an active account we’ll send you an e-mail for password recovery

Or login if you have your password back