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Each coin in a bin has a value attached to it. Each time that a coin with value p is 铿俰pped, it lands on heads with a probability p. When a coin is randomly chosen from the bin, its value is uniformly distributed on ($0,1$). Suppose that after the coin is chosen but before it is 铿俰pped, you must predict whether it will land on heads or on tails. You will win $1$if you are correct and will lose $1$otherwise.

(a)What is your expected gain if you are not told the value of the coin?

(b) Suppose now that you are allowed to inspect the coin before it is 铿俰pped, with the result of your inspection being that you learn the value of the coin. As a function of p, the value of the coin, what prediction should you make?

(c) Under the conditions of part(b), what is your expected gain?

Step by step solution

01

Given Information (part a)

What is your expected gain if you are not told the value of the coin

02

Explanation (part a)

Your expected gain if you are not told the value of the coin is$0$

03

Step 3: Final Answer (part a)

The expected gain is zero.

04

Given Information (part b)

As a function of p,the value of the coin, what prediction should you make

05

Explanation (part b)

$\text{We have function}p\text{. Predict heads if}p>\frac{1}{2}\text{.}$

06

Final Answer (part b)

As a function of p, the value of the coin, $\text{Predict heads if}p>\frac{1}{2}\text{.}$

07

Given Information (part c)

Under the conditions of part(b), what is your expected gain?

08

Explanation (part c)

The expected gain is,

$E\left[\text{gain}\right]={鈭�}_0^{1}E[\mathrm{gain}鈭�V=p]dp$

$={鈭�}_0^{\frac{1}{2}}\left[1\right(1鈭�p)鈭�1(p\left)\right]dp+{鈭�}_{\frac{1}{2}}^1\left[1\right(p)鈭�1(1鈭�p\left)\right]dp$

$={鈭�}_0^{\frac{1}{2}}(1鈭�p鈭�p)dp+{鈭�}_{\frac{1}{2}}^1(p鈭�1+p)dp$

$={鈭�}_0^{\frac{1}{2}}(1鈭�2p)dp+{鈭�}_{\frac{1}{2}}^1(2p鈭�1)dp$

$={[p鈭�\frac{2p^2}{2}]}_0^{\frac{1}{2}}{[\frac{2p^2}{2}鈭�p]}_{\frac{1}{2}}^1$

$=[\frac{1}{2}鈭�\frac{1}{4}]+[1鈭�1鈭�\frac{1}{4}+\frac{1}{2}]$

$=\left[\frac{2鈭�1}{4}\right]+\left[\frac{鈭�1+2}{4}\right]$

$=\frac{1}{4}+\frac{1}{4}$

=$\frac{2}{4}$

=$\frac{1}{2}$

09

Final Answer (part c)

The expected gain is $\frac{1}{2}$

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