Let $X$ and $Y$ be independent and identically distributed $U(0,1)$ random variables then
$$P\bigl( Y<(X-1/2)^2\bigr)$$ is?
$X$ and $Y$ are $i.i.d.$ random variables belonging to uniform distribution. Find probability of $Y$ less than a function of $X$
0
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probability
probability-distributions
uniform-distribution
2 Answers
2
Draw $y=(x-\frac{1}{2})^2$ in $XY-$plane. It is $y=x^2$ with a transition of size $\frac{1}{2}$ to the right.
The desired area can be calculated by integration. which is
$$2 \times \int_0^{\frac{1}2} x^2dx=2 \times \frac{(\frac{1}{2})^3}{3}=2 \times \frac{1}{24}= \frac{1}{12}$$
Hence the total area is $1$ the answer is $\frac{1}{12}$.
1
Since $X$ and $Y$ are independent, you know that the joint distribution of $X$ and $Y$ is the product of the marginal distributions. Since they both have $U(0,1)$-distribution, you obtain $f_{(X,Y)}(x,y) = 1$ if $(x,y) \in [0,1]^2$ and $f_{(X,Y)}(x,y) = 0$ everywhere else.
Now remember that $$P((X,Y) \in A) = \int_{A} f_{(X,Y)}(x,y) dx dy$$ for any measurable subset $A \subseteq \mathbb{R}^2$. In your case, $A = \{(x,y) \:|\: y \leq (x - \frac{1}{2})^2\}$. The integration is not difficult, so I leave that part to you.