amhailim
2022-09-23
Answered

Let X and Y be independent random variables each having the uniform density on {0,1,...,N}. Find $$P(X\ge Y)$$

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belidla5a

Answered 2022-09-24
Author has **8** answers

Using the law of the total probability (conditioning on X), we have that

$$P(X\ge Y)=\sum _{x=0}^{N}P(X\ge Y|X=x)P(X=x)$$

$$=\sum _{x=0}^{N}P(Y\le x)P(X=x)$$

$$=\sum _{x=0}^{N}\frac{x+1}{N+1}\ast \frac{1}{N+1}$$

$$=\frac{1}{(N+1{)}^{2}}\sum _{x=0}^{N}x+1$$

$$=\frac{(N+1)(N+2)}{2(N+1{)}^{2}}=\frac{N+2}{2(N+1)}$$

$$P(X\ge Y)=\sum _{x=0}^{N}P(X\ge Y|X=x)P(X=x)$$

$$=\sum _{x=0}^{N}P(Y\le x)P(X=x)$$

$$=\sum _{x=0}^{N}\frac{x+1}{N+1}\ast \frac{1}{N+1}$$

$$=\frac{1}{(N+1{)}^{2}}\sum _{x=0}^{N}x+1$$

$$=\frac{(N+1)(N+2)}{2(N+1{)}^{2}}=\frac{N+2}{2(N+1)}$$

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