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Let $X$ and $Y$ be independent random variables, taking values in the positive integers and having the same mass function $f(x)=2^{-x}$ for $x=1,2,...$ .Find $P(X\geq kY)$, for a given positive integer

I did: $\displaystyle P(X\geq kY)=P(Y\leq X/k)=\sum_{r=1}^{\infty}P(Y\leq r, X=rk)=\sum_{r=1}^{\infty}P(Y\leq r)P(X=rk)$

$P(Y\leq r)=\displaystyle\sum_{u=1}^{r}\frac{1}{2^{u}}= 1-\frac{1}{2^{r}}$

$P(X\geq kY)=\displaystyle \sum_{r=1}^{\infty} (1-\frac{1}{2^{r}})\frac{1}{2^{rk}}=\frac{2^k}{(2^k-1)(2^{k+1}-1)}$

But the solution says $P(X\geq kY)=\displaystyle\frac{2}{2^{k+1}-1}$ and it's solved by doing $\displaystyle P(X\geq kY)=\sum_{r=1}^{\infty}P(X\geq kr,Y=r)$ i think that this approach can't be so far from mine but results are different. Am i missing something? Thanks beforehand.

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When you decompose $P(X\ge kY)$ as a series of $P(Y\le r,X=rk)$, you only take into account the event that $X$ is a multiple of $k$, hence you forget the contribution of the event $[X=n]$ for every $n$ which is not a multiple of $k$. – Did Jun 2 '11 at 19:05
When you have a problem about integers, going to "fractions" (in your case, $X/k$) can be a dangerous step. It is possible to proceed without error, but I think that the probability of error increases. I have noticed this particularly in student solutions of number-theoretic problems. – André Nicolas Jun 2 '11 at 20:44
up vote 1 down vote accepted

To understand your mistake, fix $k=2$. In the case you iterate over $X=2 r$, thus missing all the odd numbers. For $Y\leq X/k$, $X$ must be conditions to be $Y\leq q$ and $X=q*r+s$ where $s$ runs from $0$ to $k-1$.

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$$ P(X \geq kY) = \sum_{y=1}^{\infty} P(X \geq kY \ | \ Y = y) P(Y = y) = \sum_{y=1}^{\infty} P(X \geq ky) 2^{-y} $$


$$ P(X \geq ky) = 1 - P(X < ky) = 1 - \sum_{r=1}^{ky - 1} 2^{-r} = 1 - 1 + (1/2)^{ky-1} = (1/2)^{ky-1} $$


$$ P(X \geq kY) = \sum_{y=1}^{\infty} 2^{-ky+1} 2^{-y} = 2\sum_{y=1}^{\infty} 2^{-y(k+1)} = 2\frac{2^{-(k+1)}}{1 - 2^{-(k+1)}} = \frac{2}{2^{k+1} - 1} $$

which is the answer your solution gave.

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