Greater Expected Outcome

Solution & Explanation

To determine which scenario yields a higher expected outcome, we must compare the expected values of two different processes:

1. Squaring a Single Number:

   • Select a single number, $X$, uniformly from the set $\{1, 2, ..., N\}$.
   • Compute the expected value of $X^2$.

2. Multiplying Two Numbers:

   • Select two numbers, $X_1$ and $X_2$, independently and uniformly from the set $\{1, 2, ..., N\}$.
   • Compute the expected value of $X_1 \times X_2$.

Mathematical Formulation:

1. Expected Value of Squaring a Single Number:

• The expected value $E(X^2)$ is calculated as: $E(X^2) = \frac{1}{N} \sum_{x=1}^{N} x^2$

• Using the formula for the sum of squares: $\sum_{x=1}^{N} x^2 = \frac{N(N+1)(2N+1)}{6}$

• Thus, $E(X^2) = \frac{1}{N} \times \frac{N(N+1)(2N+1)}{6} = \frac{(N+1)(2N+1)}{6}$

2. Expected Value of Multiplying Two Numbers:

• The expected value $E(X_1 \times X_2)$ is: $E(X_1 \times X_2) = E(X_1) \times E(X_2)$

• Since $X_1$ and $X_2$ are independently drawn from the same distribution: $E(X) = \frac{1}{N} \sum_{x=1}^{N} x = \frac{N+1}{2}$

• Thus, $E(X_1 \times X_2) = \left(\frac{N+1}{2}\right)^2$

Comparing the Expected Values:

• Variance Insight:

  • The variance of a uniform distribution is given by: $Var(X) = E(X^2) - E(X)^2$

• From the variance formula, we know: $E(X^2) = Var(X) + E(X)^2$

• Since variance is always non-negative, it follows that: $E(X^2) \geq E(X)^2$

• Therefore, $E(X^2)$, which corresponds to squaring a single number, is always greater than or equal to $E(X_1 \times X_2)$, which corresponds to multiplying two numbers independently.

Conclusion:

The expected value of squaring a single number from 1 to $N$ is greater than the expected value of multiplying two independently drawn numbers from the same range. This is due to the inherent positive variance in the distribution, which ensures that $E(X^2)$ is greater than $E(X)^2$. Thus, the first scenario yields a higher expected outcome.