0

Ms. A selects a number X randomly from the uniform distri- bution on [0,1]. Then Mr. B repeatedly, and independently, draws numbers Y1,Y2 ,.... from the uniform distribution on [0,1], until he gets a number larger than X/2, then stops. What is the expected sum of the draws by Mr. B, given X=x? This is a modification of a question I posted earlier, where you had to find the expected number of draws by Mr.B required for one success.

Please give me a hint or something for this one, I don't know where to begin.

Andy
  • 141

1 Answers1

1

Let $N$ be the index of the first time the event $\{Y_k\geq X/2\}$ occurs.

That is that $N:(Y_N> X/2)\wedge \bigwedge\limits_{k=1}^{N-1} (Y_k\leq X/2)$

So the series, $\sum_{k=1}^N Y_k$ is the sum of $(N-1)$ values of $Y_\star$ known to at most $X/2$ and $Y_N$, the first such value that is greater.

Thus since Expectation is Linear :

$$\begin{align}\mathsf E\Big(\sum_{k=1}^N Y_k\mid X=x\Big) ~=~ & \big(\mathsf E(N\mid X=x)-1\big)~\mathsf E(Y_\star\mid Y_\star\leq x/2)+\mathsf E(Y_\star\mid Y_\star\gt x/2) \end{align}$$

Now because the $\{Y_\star\}$ values are iid uniform $[0;1]$ we know $\mathsf E(Y_\star\mid Y_\star\leq x/2) ~=~ \boxed ?$ and $\mathsf E(Y_\star\mid Y_\star > x/2) = \boxed ?$.

Also the conditional distribution of $N$ given $X$ is (what?), which means that: $$\mathsf E(N\mid X=x) ~=~ \boxed?$$

Graham Kemp
  • 129,094