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There are two envelopes, each of which has a check for a Unif(0, 1) amount of money, measured in thousands of dollars. The amounts in the two envelopes are independent. You get to choose an envelope and open it, and then you can either keep that amount or switch to the other envelope and get whatever amount is in that envelope. Suppose that you use the following strategy: choose an envelope and open it. If you observe U, then stick with that envelope with probability U, and switch to the other envelope with probability 1 − U.

(a) Find the probability that you get the larger of the two amounts.

(b) Find the expected value of what you will receive.

Spack Jarrow
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2 Answers2

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Since the amounts in the envelopes are independent, the amount $U_2$ in the second envelope has distribution uniform(0,1). This is less than $u \in [0,1]$ with $u$, and less with probability $1-u$. Given that the amount in the first envelope is $u_1$, you will get the larger amount if either

  1. you switch (which happens with probability $1 - u_1$) and $U_2 > u_1$, or
  2. you don't switch (which happens with probability $u_1$) and $U_2 < u_1$.

Thus the conditional probability that you get the larger amount, given $U_1 = u_1$, is $$ P(L | U_1 = u_1) = (1-u_1)^2 + u_1^2 = 1 - 2 u_1 + 2 u_1^2$$ and the unconditional probability is $$ P(L) = \mathbb E[P(L | U_1)] = 1 - 2 \mathbb E[U_1] + 2 \mathbb E[U_1^2] = 2/3$$

Similarly, the conditional expectation of the amount $Y$ you receive, given $U_1 = u_1$, is $$ \mathbb E[V |U_1 = u_1] = (1-u_1)(1/2) + u_1 (u_1) = 1/2 - u_1/2 + u_1^2$$ and $$\mathbb E[V] = \mathbb E[\mathbb E[V|U_1]] = 1/2 - \mathbb E[U_1]/2 + \mathbb E[U_1^2] = 7/12 $$

Robert Israel
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Let $U$ denote the amount of money in the firstly chosen envelop that is opened and $V$ the amount of money in the non-chosen envelop. Then $1_{\left(0,1\right)^{2}}(u,v)$ is the PDF of $\left(U,V\right)$.

(a)

If $E$ denotes the event that the larger amount of money will eventually be received then:

$P\left(E\mid U=u,V=v\right)$ takes value $u$ if $u>v$ and takes value $1-u$ otherwise. So:

$$P\left(E\right)=\int_{0}^{1}\int_{0}^{1}P\left(E\mid U=u,V=v\right)dvdu=\int_{0}^{1}\int_{0}^{u}udvdu+\int_{0}^{1}\int_{u}^{1}\left(1-u\right)dvdu$$$$=\int_{0}^{1}u^{2}du+\int_{0}^{1}\left(1-u\right)^{2}du=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$$

(b)

If $R\left(U,V\right)$ denotes what is received then $R\left(U,V\right)=U.U+\left(1-U\right).V=U^{2}+V-UV$

So:

$$\mathbb{E}R\left(U,V\right)=\mathbb{E}\left(U^{2}+V-UV\right)=\mathbb{E}U^{2}+\mathbb{E}V-\mathbb{E}U\mathbb{E}V=\frac{1}{3}+\frac{1}{2}-\frac{1}{2}\frac{1}{2}=\frac{7}{12}$$

drhab
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  • Hi! Would you mind telling me what's wrong with the following argument? I can't find the mistake: Let $X$ be the amount received and $A$ the event of getting the larger value. Then,

    $E(X)=E(X|A)P(A)+E(X|A^c)P(A^c)=E(max(U,V))\frac{2}{3}+E(min(U,V))\frac{1}{3}=\frac{2}{3}\frac{2}{3}+\frac{1}{3}\frac{1}{3}=\frac{5}{9}$

    Thank you in advance!

     – Matías A Feb 26 '21 at 01:04
  • @MatíasA Your solution implicitly assumes that the distribution of $\min(U,V)$ is the same as the conditional distribution of $(X|A)$. Same story for $\min(U,V)$ and $(X|A^c)$. That is tempting (I remember my own struggling with that) but false. For $x\in(0,1)$ we find easily that $P(\max(U,V)\leq x)=x^2$ but things are more complicated for finding $P(X\leq x\mid A)$. It can be shown that this equals $\frac14x^3+\frac34x^2$. That gives $\mathbb E(X|A)=\frac{11}{16}$. Also we can find that $\mathbb E(X|A^c)=\frac38$. Using these values we find $\mathbb EX=\frac7{12}$. – drhab Feb 26 '21 at 10:43
  • Thank you! I would never have guessed that max$(U,V)$ and $X|A$ have different expectations. – Matías A Feb 26 '21 at 15:55