I knew the answer involved a normal distribution but fumbled the parameters for a second.
State the distribution of the difference of two independent normal variables by applying the linear combination property of normal distributions. Compute the mean and variance of the difference using the given parameters, and clearly present the final distribution.
Pro tip: Mention that independence is crucial for the variance of the difference to be the sum of variances; if they were dependent, you would need to account for covariance. Also, relate this to practical applications like A/B testing where comparing two independent groups often involves such differences.
Let X and Y be independent and identically distributed normal variables with mean μ = 1 and variance σ² = 3. So X ~ N(1, 3) and Y ~ N(1, 3).
We want the distribution of D = X - Y. Since X and Y are independent, D is a linear combination of independent normal variables.
A linear combination of independent normal variables is normally distributed. Thus, D follows a normal distribution.
The mean of D is E[X] - E[Y] = 1 - 1 = 0. The variance is Var(X) + Var(Y) = 3 + 3 = 6 (since independent, variances add even for subtraction).
Therefore, D ~ N(0, 6).
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