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Math Help - Uniformly Distributed Random variables

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    Uniformly Distributed Random variables

    Hey guys, another one of these crazy probability questions...

    Let X and Y be independent, each uniformly distributed on {1, 2, ... , n}.

    Find P(X + Y = k) for 2 <= k <= 2n.

    This is the last of a series of problems, all of the previous ones which were easily solved by drawing up a grid of possible choices. Is there a way to figure this out WITH OUT a grid?

    A gentle shove in the right direction would be helpful! Thanks!
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    Hello,
    Quote Originally Posted by Dfowj View Post
    Hey guys, another one of these crazy probability questions...

    Let X and Y be independent, each uniformly distributed on {1, 2, ... , n}.

    Find P(X + Y = k) for 2 <= k <= 2n.

    This is the last of a series of problems, all of the previous ones which were easily solved by drawing up a grid of possible choices. Is there a way to figure this out WITH OUT a grid?

    A gentle shove in the right direction would be helpful! Thanks!
    think of it as events.

    Let's consider the possible values for Y, in the event {X+Y=k}
    Y can take any value between {1,2,...,n}

    So you can divide the probability space into events :
    \{X+Y=k\} = \bigcup_{j=1}^n \{X+Y=k  ~ and ~ Y=j\}

    So this gives \mathbb{P}(X+Y=k)=\sum_{j=1}^n \mathbb{P}(X+Y=k ~,~ Y=j), since the events are disjoint (Y cannot take two values at a same time)

    ---> \mathbb{P}(X+Y=k)=\sum_{j=1}^n \mathbb{P}(X+j=k ~,~ Y=j)=\sum_{j=1}^n \mathbb{P}(X=k-j ~,~ Y=j)

    =\sum_{j=1}^n \mathbb{P}(X=k-j)\mathbb{P}(Y=j) because X and Y are independent.

    Does it look clear to you ?
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