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My professor did this problem out of our text book and he didn't exactly show us how he did it (he skipped showing the steps in the middle). He got the answer $(1-x)^2$.

If we let $X$ and $Y$ be two independent uniform $(0,1)$ random variables and let $M$ be the minimum of $X$ and $Y$, $0\lt x \lt 1$. How would we represent the event $(M\ge x)$ as the region in the plane and find $P(M\ge x)$ in the area of this region?

How would you go about doing this?

PS I added the homework tag because it could have been a homework problem, however it's not homework. I'm just looking for more people to see my question.

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The event $M \geq x$ is the intersection of the two events $X \geq x$ and $Y\geq x$. In the plane, this means that the location of $(X,Y)$ must be to the upper-right of $(x,x)$, so $(X,Y)$ must belong to the square $[x,1)\times [x,1)$. This square has area $(1-x)^2$, which is the desired probability.

In other words: $\begin{align*} P(M\geq X) &= P(X\geq x\text{ and }Y\geq x)\\ &= P(X\geq x)P(Y\geq x)\text{ since $X$ and $Y$ are independent}\\ &= m([x,1))m([x,1))\text{ since $X$ and $Y$ are uniform $(0,1)$-random variables}\\ &= (1-x)(1-x)\text{ since the measure of an interval is its length.} \end{align*}$

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    The CDF is the probability that $M$ is any value up to $x$: P(M < x) = 1 - P(M\geq x) = 1 - (1-x)^2. The PDF is the derivative of the CDF: $d(1-(1-x)^2)/dx = 0 - 2(1-x)^1 (-1) = 2(1-x)$.2012-11-20
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The probability that $M\ge x$ equals the probability that both $X$ and $Y$ are $\ge x$. Since each of these is $1-x$ and they are independent, the result follows.