# uniform distribution probablity [closed]

Let $X$ and $Y$ be independent random variables each having the uniform distribution on $\{0, 1, ... , N\}$.

Find

i) $P(\min(X,Y))$

ii) $P(\max(X,Y))$

iii) $P(|Y-X|)$ .

I always get confused with this kind of problem of minimum and maximum in probability. Can any general method be applied here?

## closed as off-topic by Did, flawr, Em., user26857, Stefan MeskenFeb 28 '16 at 21:19

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## 2 Answers

First, fix a certain number $z$ between $0$ and $N$. Note that $\min(X,Y)\geq z$ only if both $X$ and $Y$ are larger than $z$. Since $X$ and $Y$ are independent, their joint distribution factorizes (as per @flawr answer above) $$P_{X,Y}(X=x,Y=y)=P_X(X=x)P_Y(Y=y)=\frac{1}{(N+1)^2}\ .$$ The general procedure is now summing over all possible values your random variables can take (weighted with their joint probabilities), such that a certain condition is verified (namely $X$ and $Y$ both larger than $z$) $$P(\min(X,Y)\geq z)=\sum_{x=0}^{N}\sum_{y=0}^{N}P_{X,Y}(X=x,Y=y)\Theta(x-z)\Theta(y-z)\ ,$$ where $\Theta(t)=1$ for $t\geq 0$ and $=0$ otherwise. Therefore $$P(\min(X,Y)\geq z)=\frac{1}{(N+1)^2}\sum_{x=z}^{N}\sum_{y=z}^{N}1=\frac{(1+N-z)^2}{(1+N)^2}\ .$$ [The initial summation limits are $(0,N)$ as these are all possible values in the alphabets that $X$ and $Y$ can take. Due to the Theta constraints, the summation range gets truncated, and only goes from $z$ to $N$ (all the other summands are zero). To compute the sum of $1$ over a certain range, well, I don't think I have to explain this.] For example if $z=0$, the probability that $\min(X,Y)\geq 0$ is $1$ (as the minimum of two numbers between $0$ and $N$ is surely larger or equal to zero). If $z=N$, this probability is $1/(N+1)^2$, because out of the $(N+1)^2$ possible pairs of values $(X,Y)$, only the pair $(X=N,Y=N)$ would match the requirement $P(\min(X,Y)\geq N)$. Can you take it from here?

Edit. From $P(\min(X,Y)\geq z)$ we can compute the probability $P(\min(X,Y)= z)$ as $$P(\min(X,Y)= z)=P(\min(X,Y)\geq z)-P(\min(X,Y)\geq z+1)=$$ $$\frac{(1+N-z)^2}{(1+N)^2}-\frac{(1+N-z-1)^2}{(1+N)^2}=\frac{2 N-2 z+1}{(N+1)^2}\ ,$$ as per textbook solution given in the comment.

• how to do summation and set limit for summation also answer in book is given as 2(N-z)+1/(N+1)^2 – maths student Feb 28 '16 at 10:20
• same trick can be used for maximum or not – maths student Feb 28 '16 at 10:43
• Sure. Try and write back if you need more help. – Pierpaolo Vivo Feb 28 '16 at 12:57

There is one "trick" you can apply when dealing with minima and maxima: Note that

$$P(\max(X,Y) < x) = P(X < x \text{ and } Y < x)$$

• But how we proceed after that? – maths student Feb 28 '16 at 9:54
• As $X$ and $Y$ are independend, you can factorize $P(X < x \text{ and } Y < x ) = P(X < x) P(Y < x)$ – flawr Feb 28 '16 at 9:57
• but how to calculate probablity this is the main thing that I don"t understand – maths student Feb 28 '16 at 10:00
• You already confirmed the formula that @PierpaoloVivo wrote in his comment. – flawr Feb 28 '16 at 10:02
• X=Y case is also there\ – maths student Feb 28 '16 at 10:03