Find the probability of $\max[X,Y]<\frac{1}{2}$?

Question

$X$ and $Y$ are uniformly distributed random variable on $[-1,1]$. Find the probability of $\max[X,Y]<\frac{1}{2}$.

I calculated it using graphically that the region where the where $[X,Y]$ is less than half divided by the total area i.e. $9/4*1/4=9/16$ but i have a doubt that this approach of mine has included the points where $\max[X,Y]=1/2$.

Answer 1

Assuming they are independent, let $M = \max\{X,Y\}$. Then, \begin{align*} P(M\leq 1/2) &= P(X\leq 1/2, Y\leq 1/2)\tag 1\\ &= P(X\leq 1/2)P(Y\leq 1/2)\tag 2\\ &=\left(\frac{1/2-(-1)}{1-(-1)}\right)\left(\frac{1/2-(-1)}{1-(-1)}\right)\tag 3\\ &=\left(\frac{3/2}{2}\right)^2\\ &=\frac{9}{16} \end{align*} where:

• $(1)$ is true since if I want the max to be less than $1/2$, then both $X$ and $Y$ have to be less than $1/2$

• $(2)$ is true by independence

• (3) is true since the probability we want geometrically is $$\frac{\text{distance from $-1$ to $1/2$}}{\text{distance from $-1$ to $1$}}.$$

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You stated "the where $[X,Y]$ is less than half divided by the total area", but this is $$\frac{\text{Area less than half}}{\text{Total area}} = \frac{\frac{1}{2}(1/2-(-1)}{1} = \frac{3}{4}$$ but this only accounts for one random variable $X$ or $Y$. It needs to be the case that $X$ and $Y$ follow this, so you must square: $$P(M\leq 1/2) = P(X\leq 1/2)P(Y\leq 1/2) = \frac{3}{4}\cdot\frac{3}{4} = \frac{9}{16}.$$

Answer 2

I calculated using this way the blue shaded area is$\frac{9}{4}$and the total area is $4$ . Hence the probability will be $\frac{9}{4}*\frac{1}{4}$