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Question: I pick a constant $a$ between 0 and 1. You repeatedly draw random numbers as follows. The first random number is drawn uniformly from (0,1); the second random number is drawn uniformly between 0 and the first number. Each subsequent random number is drawn uniformly between 0 and the previous number. You stop until you obtained a random number that is below a. Find the expected number of drawings as a function of a. (Hint: condition on the first number)

If $X_1, X_2,$ ... is the sequence of random number drawings

and $M_a = \min( n \geq 1: X_n < a)$ is the number of minimum drawings until we draw a number less than a.

I know that $P(M_a=1) = P(X_1 < a) = a$.

My professor said that we should suppose

$P(M_a=k) = \frac{e^{-(-\log a)(-\log a)^{k-1}}}{(k-1)!}$

Does anyone mind explaining why we can assume this?

Also, he mentioned that

$$P(M_a=k+1) =\int_0^1 E(M_a=k+1 | X1=x) \,dx$$

Can someone explain the logic behind this integral?

Thank you!

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    are you sure about the formula for $P(M_a = k)$? – 2017-01-11
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    *Watch out for the typos in your question...* Conditionally on $X_1$, the sample $(X_2,\ldots,X_{k+1})$ is distributed like $(X_1X'_1,\ldots,X_1X'_k)$ where $(X'_1,\ldots,X'_k)$ is independent of $X_1$ and distributed like $(X_1,\ldots,X_k)$. Thus, $M_a=1$ on $[X_1a]$. This implies that $P(M_a=1)=a$ and, for every positive $k$, $$P(M_a=k+1)=\int_a^1P(M_{a/x}=k)dx=a\int_a^1P(M_y=k)y^{-2}dy$$ Assuming that $$P(M_a=k)=\frac{1}{(k-1)!}a(-\log a)^{k-1}$$ for every $a$, for some positive $k$, one gets ... – 2017-01-11
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    $$P(M_a=k+1)=\frac{a}{(k-1)!}\int_a^1(-\log y)^{k-1}y^{-1}dy=\frac{1}{k!}a(-\log a)^k$$ which proves the desired property for every $k$ since it holds for $k=1$ (sorry but I lack the courage to compare this with your suggestion). Finally, $$P(M_a=k+1)=E(P(M_a=k+1\mid X_1))=\int_0^1P(M_a=k+1\mid X_1=x)f_{X_1}(x)dx$$ is simply Bayes formula. – 2017-01-11
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    @Did Why don't you transform your remarkable comment into an answer ? – 2017-01-11
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    In order to bring back your formula in conformity with the exact formula as given by @Did, have you remarked that $e^{log(a)}$ is plainly ... $a$... under the condition to lower $(-log(a))^{k-1}$ to the "ground floor"... – 2017-01-11

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