Let X, Y be independent RVs. Calculate the following probabilities Let $X, Y$ be independent random variables with positive integers values, with distribution
$$ P(X=i)=P(Y =i)= \frac{1}{2^i},i∈N^∗$$
Find the following probabilities.
(i) $P(\max(X, Y ) ≥ i)$ 
(ii) $P(X = Y)$
(iii) $P(X > Y)$
 A: Hints. Recall that independence gives you that 
$$ \def\P#1{\mathbf P\left[#1\right]}\P{X \in A, Y \in B} = \P{X\in A}\P{Y\in B} $$
for every $A,B \subseteq \def\N{\mathbf N}\N^*$. 
(i) Note that we have 
$$ \P{\max(X,Y)\ge i} = 1 - \P{\max(X,Y)<i} = 1 - \P{X < i, Y < i} $$
and use independence.
(ii) Here the law of total probiability will help you, conditioning on $Y$ gives 
$$ \P{X=Y} = \sum_{i\in \N^*} \P{X=Y \mid Y = i}\P{Y=i} 
     = \sum_{i\in \N^*} \P{X = i\mid Y = i}\P{Y=i}
$$
Now independence will help you again.
(iii) Use symmetry in $X$ and $Y$ of the problem to conclude $\P{X<Y}=\P{Y<X}$. Now use (ii). 
A: Try to go back to more elementary problems when facing this kind of questions.


*

*$max(X,Y) \geq i$ is equivalent to $X \geq i$ or $Y \geq i$.


So $P(max(X,Y) \geq i) = P(X \geq i) + P(Y \geq i) - P(X \geq i, Y \geq i) = 2 \sum_{j=i}^{\infty} \frac{1}{2^i} - \left( \sum_{j \geq i} \frac{1}{2^j} \right)^2$


*Don't forget that $X$ and $Y$ are independent: $P(X = Y) = \sum_i P(X=i, Y = i) = \sum_i P(X = i) P(y = i) = \sum_i \frac{1}{2^{2i}}$

*Use conditional probabilities: $P(X > Y) = \sum_i P(X > Y | Y = i) P(Y=i) = \sum_i P(X > i) \frac{1}{2^i} = \sum_i \frac{1}{2^i} \sum_{j > i} \frac{1}{2^j}$
I hope my advices will help you with future problems.
