Calculating the Expected Value in a Gaussian Sum

Solution & Explanation

Problem Statement: We are given two Gaussian random variables, $X_1$ and $X_2$, with means $m_1$ and $m_2$, and variances $v_1$ and $v_2$ respectively. We know that the sum $X_1 + X_2 = n$. We need to determine the expected value of $X_2$.

Understanding the Problem:

• The sum $X_1 + X_2 = n$ implies a linear relationship between $X_1$ and $X_2$.
• The expected value of a Gaussian random variable is its mean.

Key Concept: Linearity of Expectation

• The linearity of expectation states that for any two random variables $X$ and $Y$, $E[X + Y] = E[X] + E[Y]$. This property holds regardless of whether the random variables are independent.

Solution:

1. Setup the Equation:

   • We know $X_1 + X_2 = n$.
   • By the linearity of expectation, we have: $E[X_1 + X_2] = E[X_1] + E[X_2] = n$
   • Substitute the known expected values (means) of $X_1$ and $X_2$: $m_1 + E[X_2] = n$

2. Solve for $E[X_2]$:

   • Rearranging the equation gives: $E[X_2] = n - m_1$

Explanation:

• The solution utilizes the linearity of expectation, which simplifies the calculation.
• The fact that $X_1$ and $X_2$ are Gaussian is not directly relevant to finding the expected value of $X_2$ given $X_1 + X_2 = n$. This is because the linearity of expectation applies to all random variables, not just Gaussian ones.
• The dependency introduced by $X_1 + X_2 = n$ is handled by the rearrangement of the linear expectation equation.

Conclusion:

• The expected value of $X_2$ given the sum $X_1 + X_2 = n$ is $n - m_1$. This result is derived purely from the properties of the expectation operator and does not depend on the Gaussian nature of the variables.

Note:

• This type of problem is useful to test understanding of the linearity of expectation and familiarity with Gaussian properties, although the Gaussian property is not used here.