X and Y are independent uniform random variables distributed in [-1,1], how do I find $P[ X^2 < \frac{1}{2}, |Y| < \frac{1}{2}]$?

Question

• Since $X$ and $Y$ are uniformly distributed in $[-1,1]$:

$f_Y(y) = f_X(x) = \frac{1}{1-(-1)} = \frac{1}{2}$ over $-1 \leq y \leq 1$ and $-1 \leq x \leq 1$ respectively, and $0$ everywhere else for each of the RVs.

• Then the only place $Y < \frac{1}{2} = 1 - \int_{-1}^{1} f_Y(y) = 0$

• It seems intuitive to me that given the above, regardless of the first constraint $|X| < \frac{1}{2}$ should then be 0. Since $0 \cap S = 0$

However the answer in my solution manual indicates that:

$P[|X|<\frac{1}{2}]P[Y<\frac{1}{2}] = \frac{1}{2}\frac{1}{\sqrt 2}$, and I'm not sure how it arrives there...

What am I doing incorrectly here?

Answer 1

Since $X$ and $Y$ are independent, we have $$\Pr[(X^2 \le 1/2) \cap (|Y| \le 1/2)] = \Pr[X^2 \le 1/2]\Pr[Y \le 1/2].$$ Since they are uniform on $[-1,1]$, we then have $$\Pr[X^2 \le 1/2] = \Pr[-1/\sqrt{2} \le X \le 1/\sqrt{2}] = \frac{1/\sqrt{2} - (-1/\sqrt{2})}{1 - (-1)} = \frac{1}{\sqrt{2}},$$ and $$\Pr[|Y| \le 1/2] = \Pr[-1/2 \le Y \le 1/2] = \frac{(1/2)-(-1/2)}{1-(-1)} = \frac{1}{2},$$ and the claimed result follows. In both cases, we used the fact that if a random variable $U$ is uniformly distributed on $[a,b]$ with $a < b$, we have $$\Pr[c < U < d] = \frac{d-c}{b-a}, \quad a < c < d < b.$$

Answer 2

Hint:

1. $X^2< \frac12\iff |X|<\frac1{\sqrt2}$.
2. If $X$ and $Y$ are independent, then $|X|$ and $|Y|$ are also independent.
3. Complete the sentence: if $A$ and $B$ are independent, then $P(A, B) = ...$

Answer 3

Notice that because of independence: \begin{equation} P\left(X^2 < \frac{1}{2}, |Y| < \frac{1}{2} \right) = P\left(X^2 < \frac{1}{2}\right) P\left(|Y| < \frac{1}{2} \right) \end{equation}

Analyzing the $X$ term: \begin{equation} P\left(X^2 < \frac{1}{2}\right) = P \left( -\frac{1}{\sqrt{2}} < X < \frac{1}{\sqrt{2}} \right) = \frac{1}{\sqrt{2}} \end{equation}

While the $Y$ term: \begin{equation} P\left(|Y| < \frac{1}{2} \right) = P\left(\frac{1}{2} <Y < \frac{1}{2} \right) = \frac{1}{2} \end{equation}

Substituting both results in the first equation: \begin{equation} P\left(X^2 < \frac{1}{2}, |Y| < \frac{1}{2} \right) = \frac{1}{2}\frac{1}{\sqrt{2}} \end{equation}