To show that $E(X_1 + X_2 + \dots + X_n) = E(X_1) + E(X_2) + \dots + E(X_n)$, we use the linearity of expectation.

Linearity of Expectation:

The expectation operator $E$ is linear, meaning: $E(aX + bY) = aE(X) + bE(Y)$ where $a$ and $b$ are constants, and $X$ and $Y$ are random variables.

Proof:

1. Consider the sum $S = X_1 + X_2 + \dots + X_n$: The expectation of $S$ is defined as: $E(S) = E(X_1 + X_2 + \dots + X_n).$

2. Apply the linearity of expectation: Using linearity, the expectation of a sum of random variables is the sum of their expectations: $E(X_1 + X_2 + \dots + X_n) = E(X_1) + E(X_2) + \dots + E(X_n).$

3. No assumptions on dependence or independence: This property of expectation holds regardless of whether the random variables $X_1, X_2, \dots, X_n$ are independent or dependent.

Conclusion:

We have shown that: $E(X_1 + X_2 + \dots + X_n) = E(X_1) + E(X_2) + \dots + E(X_n).$

This result relies solely on the linearity of the expectation operator, making it a fundamental and general property.

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Would you like a worked-out numerical example to illustrate this property?

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Five Related Questions:

1. How does the linearity of expectation extend to infinite sums of random variables?
2. Does the linearity of expectation depend on the random variables being independent?
3. How is the linearity of expectation applied in variance computations?
4. Can you provide examples where linearity of expectation is used in probability problems?
5. How does linearity of expectation work for random variables with different distributions?

Tip:

When working with expectations, always remember that linearity holds regardless of independence or the distribution of random variables.