Let $X$ and $Y$ be independent and identically distributed random variables with probability mass function $p(n)=\frac{1}{2^{n}}$. Find $P(X \geq 2Y)$.

My attempt: \begin{align} P(X \geq 2Y) & = \sum\limits_{y=1}^{\infty}\sum\limits_{x=2y}^{\infty}P(X) \\ & = \sum\limits_{y=1}^{\infty} \left( \frac{1}{2^{2y}}+\frac{1}{2^{2y+1}}+\frac{1}{2^{2y+2}} + \ldots \right) \\ & = \sum\limits_{y=1}^{\infty}\frac{1}{2^{2y}} \left( 2 \right) \\ & = \frac{2}{3}.\end{align}

$\frac{2}{3}$ however does not match the answer at the back. What am I doing wrong? Help!

(The answer is given to be in the interval $[0.27,0.3]$.)


We have $$\mathbb{P}(X \geq 2Y) = \sum_{y \geq 1} \mathbb{P}(X \geq 2Y, Y=y) = \sum_{y \geq 1} \sum_{x \geq 2y} \mathbb{P}(X=x, Y=y) = \sum_{y \geq 1} \sum_{x \geq 2y} \frac{1}{2^x} \frac{1}{2^y}.$$


$$\mathbb{P}(X \geq 2Y) = \sum_{y \geq 1} \frac{1}{2^y} \frac{2}{2^{2y}} = 2 \sum_{y \geq 1} \frac{1}{2^{3y}}= \frac{2}{7}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.