Given that X and Y are independent random variables with uniform distributions and mean 0 and standard deviation 1, what is the probability of 2X being greater than Y?

Question Analysis

This is a probability question involving independent random variables with uniform distributions. Specifically, we need to determine the probability that the value of (2X) is greater than (Y), given that both (X) and (Y) are independent and have uniform distributions with a mean of 0 and a standard deviation of 1.

Key points to consider:

• Independence: Since (X) and (Y) are independent, their joint distribution is the product of their marginal distributions.
• Uniform Distribution: Typically, uniform distributions are defined over a specific interval. Here, the mention of mean 0 and standard deviation 1 suggests consideration of a standard normal distribution instead, as uniform distribution inherently doesn't have these parameters as described.
• Transformation: The problem involves a transformation (2X), meaning we need to account for how this affects the distribution and compare it with (Y).

Answer

To solve the problem, let's assume that both (X) and (Y) are actually standard normal variables since the description of uniform distribution with mean 0 and standard deviation 1 fits a standard normal distribution.

1. Transformation of Variables:

   • If (X) is a standard normal variable, (2X) is also normally distributed with mean 0 and standard deviation 2 (since multiplying a normal variable by a constant scales the standard deviation by that constant).

2. Distribution of (2X - Y):

   • Since (X) and (Y) are independent, the distribution of (2X - Y) is also a normal distribution.
   • The mean of (2X - Y) is 0 (mean of (2X)) - 0 (mean of (Y)) = 0.
   • The variance of (2X - Y) is ( (2^2 \cdot 1^2) + (1^2 \cdot 1^2) = 4 + 1 = 5 ). Therefore, the standard deviation is (\sqrt{5}).

3. Probability Calculation:

   • We are interested in (P(2X > Y)), which is equivalent to (P(2X - Y > 0)).
   • Since (2X - Y) is normally distributed with mean 0 and standard deviation (\sqrt{5}), we find this probability using the standard normal distribution:
   • (P(2X - Y > 0) = P\left(Z > 0\right)), where (Z \sim N(0, 1)).
   • From standard normal distribution tables, (P(Z > 0) = 0.5).

Conclusion: The probability that (2X) is greater than (Y) is (\mathbf{0.5}).