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Ms. A selects a number X randomly from the uniform distribution on $[0, 1]$. Then Mr. B repeatedly, and independently, draws numbers $Y_1$,$Y_2$,.... from the uniform distribution on $[0,1]$ until he gets a number larger than $X/2$, then stops. The expected number of draws that Mr. B makes equals (a) 2ln2 (b) ln 2 (c) 2/e (d) 6/e

I tried reaching a solution through integrating from X/2 to 1 to find the probability that a random variable Y would achieve a value greater than X/2. I then defined an RV Z which is equal to the maximum of all Ys picked, and imposed the condition that Z > X/2. But I got a hunch that my method was wrong and stopped midway, I cant come to think of any method either. A solution that I saw uses conditional expectation and the law of iterated expectations, but I'm finding it difficult to grasp in one stretch. This is the link to the solution that I saw,

https://sites.google.com/a/econschool.in/econschool/stuff-of-interest/anotherpost/dse-2014-q22

How do I solve this question, can somebody please help me out with a hint ?

Meera Unni
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2 Answers2

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Hint:

If $N$ denotes the number of draws needed by Mr. B then $N$ follows geometric distribution with "random parameter" $1-X/2$.

That gives: $$\mathbb EN=\mathbb E\left[\mathbb E[N\mid X]\right]=\mathbb E \left[\frac{1}{1-X/2}\right]$$

drhab
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  • How can we define N, a discrete geometric RV on something that follows a continuous distribution. Could you please give some insight, thanks a lot. I guess I've understood a lot more than before. – Meera Unni Feb 26 '18 at 09:26
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    Actually $N$ is not a random variable with geometric distribution. Though for every fixed $x\in[0,1]$ random variable $N$ has a geometric distribution with parameter $1-x/2$ under condition $X=x$. This is enough to find $\mathbb EN$ and it is not necessary to dive into the (unconditional) distribution of $N$. If you still want to then to be found is $P(N=k)=\int_0^1 P(N=k\mid X=x)dx=\int_0^1(x/2)^{k-1}(1-x/2)dx$. Doing so you will find that $N$ taken unconditionally is not a geometric random variable. – drhab Feb 26 '18 at 09:44
  • Thanks a lot drhab! – Meera Unni Mar 01 '18 at 10:38
  • You are very welcome. – drhab Mar 01 '18 at 13:15
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For a fixed $x\in [0,1]$, let $w$ be the expected number of rounds until $y > {\large{\frac{x}{2}}}$.

Then we have the equation $$w=\left(1-{\small{\frac{x}{2}}}\right)(1) + \left({\small{\frac{x}{2}}}\right)(1+w)$$ which yields $$w=\frac{2}{2-x}$$ hence, given that $x$ is uniformly distributed on $[0,1]$, the expected number of rounds is $$\int_{0}^{1}\frac{2}{2-x}\;dx = 2\ln(2)$$

quasi
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