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It doesn't really make sense to. First I'll say the problem: Let $X$ and $Y$ be two independent uniform $(0,1)$ random variables. Let $M$ be the smaller of $X$ and $Y$. Let $0\lt x\lt 1)$. Represent the event $(M\ge x)$ as the region in the plane, and find $P(M\ge x)$ as the area of this region.

So now the answer says: In terms of $X$ and $Y$, $P(M\ge x)=P(X\ge x,Y\ge x)$. As a region in the plane this is the set $\{(X,Y)|x\le X\le 1,x\le Y\le 1\}$ and the probability is $(1-x)^2$.

How is this found? I don't understand why both $X$ and $Y$ have a lower bound of $x$.

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If the minimum of two numbers has value $x$ or more, is it reasonable to assume that both numbers have value $x$ or more? If not, can you give an example of two numbers such that their minimum is $x$ or more but at least one has value smaller than $x$? As far as the probability is concerned, remember that if $X$ and $Y$ are independent, then $P\{X\in A, Y\in B\}= P\{X\in A\}P\{Y\in B\}$ for independent random variables. – Dilip Sarwate Dec 19 '12 at 15:32
up vote 2 down vote accepted

The joint density function of $X$ and $Y$ is, by independence, the product of the individual densities.

Let $S$ be the square with corners $(0,0)$, $(1,0)$, $(1,1)$, and $(0,1)$. Our joint density is $1$ in the interior of the square, and $0$ on the boundary of the square, and outside the square. It will be useful to draw the square.

So the joint density function "lives" inside the square, and is constant there. If $R$ is any region of the plane, then the probability that $(X,Y)\in R$ is just the area of the part of $R$ that is inside the square.

Now let's look at our random variable $M$, the minimum of $X$ and $Y$. We want to find the probability that $M\ge t$. Note that we have changed names slightly: the letter $x$ is perhaps too closely associated with the random variable $X$, so it may cause confusion.

We will be happy if the minimum of $X$ and $Y$ is $\ge t$. What is the probability that we will be happy?

First we deal with the easy stuff. Let $t\le 0$. Then for sure $X\ge t$ and $Y\ge t$, so for sure we will be happy. In symbols, if $t\le 0$, then $\Pr(M\ge t)=1$. Now let $t\ge 1$. Then for sure $X\le t$ and $Y\le t$, so $\Pr(M\ge t)=0$. Next we deal with the interesting part, where $0\lt t \lt 1$.

The minimum of $X$ and $Y$ is $\ge t$ precisely if both $X$ and $Y$ are $\ge t$.

Where is $X\ge t$? In the part of the plane that has $x$-coordinate $\ge t$. Draw the vertical line $x=t$. Then $X\ge t$ on that line, and to the right of it.

Where is $Y\ge t$? In the part of the plane that has $y\ge t$. Draw the horizontal line $y=t$. Then $Y\ge t$ on that line and above it.

Note that the lines $x=t$ and $y=t$ meet at $(t,t)$. We will be happy if both $X$ and $Y$ are $\ge t$. So we will be happy if $(X,Y)$ lands in the part of the world that is above and to the right of the point $(t,t)$.

But our probability density function lives inside the square. The part of the square that is above and to the right of the point $(t,t)$ is a square with sides $1-t$. This square has area $(1-t)^2$. so if $0\lt t\lt 1$, then $\Pr(M\ge t)=(1-t)^2$.

Remark: The question, as it was put, was asking you to think about the geometry of the situation. At the formula level, we can do the problem in a couple of lines. We have $M\ge t$ iff $X\ge t$ and $Y\ge t$. But $\Pr(X\ge t)=1-t$, and $\Pr(Y\ge t)=1-t$, so by independence $\Pr(M\ge t)=(1-t)^2$.

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+1 for the detailed explanation. It takes longer to write down the explanation than it does to draw a sketch of two squares with opposite vertices $(0,0)$ and $(1,1)$, and $(t,t)$ and $(1,1)$ respectively, and the answer falls out by inspection. Unfortunately, it is very hard to persuade students to draw a diagram; maybe it is drilled into them in calculus courses that sketching the graph of a function is the wrong way to approach a problem because it is easy to be led astray by not-completely-accurate sketches. – Dilip Sarwate Dec 20 '12 at 15:55

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