Independent Uniform Variables

Solution & Explanation

To solve the problem of determining the probability that $2X > Y$ where $X$ and $Y$ are independent uniformly distributed random variables with mean 0 and standard deviation 1, we need to first understand the properties of these distributions.

Uniform Distribution Basics

For a uniform distribution over an interval $[a, b]$:

• Mean (Expected Value): $E[X] = \frac{a+b}{2}$
• Variance: $Var(X) = \frac{(b-a)^2}{12}$
• Standard Deviation: $\sigma = \sqrt{Var(X)}$

Given Conditions

1. Mean: $E[X] = 0$
2. Standard Deviation: $\sigma = 1$

From the mean condition $\frac{a+b}{2} = 0$, it follows that $b = -a$.

Substituting into the variance formula: $Var(X) = \frac{(b-a)^2}{12} = 1$ $\Rightarrow \frac{(2a)^2}{12} = 1$ $\Rightarrow \frac{4a^2}{12} = 1$ $\Rightarrow a^2 = 3$ $\Rightarrow a = \sqrt{3}$

Thus, $X$ and $Y$ are uniformly distributed over $[-\sqrt{3}, \sqrt{3}]$.

Probability Calculation

We need to calculate $P(2X > Y)$. Since $X$ and $Y$ are independent, their joint distribution is the product of their marginals.

Joint PDF: $f_{X,Y}(x,y) = \frac{1}{(2\sqrt{3})^2} = \frac{1}{12} \quad \text{for} \quad x, y \in [-\sqrt{3}, \sqrt{3}]$

Probability Expression: $P(2X > Y) = \int_{-\sqrt{3}}^{\sqrt{3}} \int_{-\sqrt{3}}^{2x} f_{X,Y}(x,y) \, dy \, dx$

Integration: $P(2X > Y) = \int_{-\sqrt{3}}^{\sqrt{3}} \int_{-\sqrt{3}}^{2x} \frac{1}{12} \, dy \, dx$ $= \frac{1}{12} \int_{-\sqrt{3}}^{\sqrt{3}} (2x + \sqrt{3}) \, dx$

Evaluating the Integral: $= \frac{1}{12} \left[ x^2 + \sqrt{3}x \right]_{-\sqrt{3}}^{\sqrt{3}}$ $= \frac{1}{12} \left( (3 + 3\sqrt{3}) - (-3 - 3\sqrt{3}) \right)$ $= \frac{1}{12} (6 + 6\sqrt{3})$ $= \frac{1}{12} \times 12$ $= \frac{1}{2}$

Conclusion

The probability that $2X > Y$ is $0.5$. This result is intuitive due to the symmetry and independence of the distributions of $X$ and $Y$. Essentially, the transformation $2X$ still maintains a symmetric property around zero, leading to equal chances of $2X$ being greater or less than $Y$.