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A weighted average involves combining different values by multiplying each by a factor, known as "weight", and then summing them up. Imagine you have two sample means, \( \bar{X}_1 \) and \( \bar{X}_2 \), from independent random samples. You create a new statistic, \( \bar{X} = a \bar{X}_1 + (1-a) \bar{X}_2 \), where \( a \) is a number between 0 and 1. This number \( a \) is the weight assigned to \( \bar{X}_1 \), while \( 1-a \) is assigned to \( \bar{X}_2 \). The idea behind assigning weights is to control the influence each sample mean has on the final result, \( \bar{X} \), ensuring it's closer to an accurate estimation of the population mean, \( \mu \). These weights ensure that each sample contributes proportionately based on its variance, helping make \( \bar{X} \) an unbiased estimator for \( \mu \), as seen through \( \mathbb{E}(\bar{X}) = \mu \) when choosing weights properly.

The concept of variance reduction is crucial when trying to achieve reliable estimates. In statistics, variance measures how spread out values are around the mean. Lower variance in an estimator indicates more precision. When looking at our weighted average \( \bar{X} = a \bar{X}_1 + (1-a)\bar{X}_2 \), we want to find the "optimal" weight, \( a \), that minimizes the variance, hence increasing the estimator's accuracy.
For independent samples, the variance can be expressed as \( \text{Var}(\bar{X}) = a^2 \frac{\sigma^2}{n_1} + (1-a)^2 \frac{\sigma^2}{n_2} \). The task is to find the weight \( a \) that makes this variance as small as possible. By using calculus, we find that setting \( a = \frac{n_1}{n_1 + n_2} \) optimizes the variance reduction, leading to the smallest possible variance for \( \bar{X} \). This procedure helps in forming a more precise and credible estimator of the mean.

Independent random samples refer to samples drawn from a population in such a way that the selection of one sample does not influence or affect the selection of another. Each sample stands alone, with no overlap or connection. In statistical estimation, this independence is a key condition when combining sample data, ensuring that their respective variances are simply additive.
In our context, \( \bar{X}_1 \) and \( \bar{X}_2 \) are means from independent samples. This independence allows us to calculate the variance of the combined estimator \( \bar{X} \) as \( \text{Var}(\bar{X}) = a^2 \text{Var}(\bar{X}_1) + (1-a)^2 \text{Var}(\bar{X}_2) \). Because of their independence, cross-products terms (which could occur if they were dependent) do not appear in the variance formula, simplifying the variance expression. This simplification ultimately aids in deriving the precise value of \( a \) necessary for minimizing the variance, emphasizing the power of independent sampling in statistical analysis.