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If we a sample of $n$ values from a given population and if $X$ is the variable of the sample, then the mean of $X$ is just $\dfrac{ \sum x }{n}$

Now, suppose $X$ is random variable. For concreteness, let us take $X$ to be the number of heads in a toss of a coin. Now, $X$ can be 0 or 1. Therefore, the mean in this case is $\dfrac{0+1}{2} = \dfrac{1}{2}$ and the formula coincides with the above, the one for samples.

In general, for random variable $X$, we know the mean is $\sum x P(X=x)$.

But, how is this different from the mean for samples? What is the motivation for this defition?

James
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2 Answers2

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If a sample of n values are given and each weight of sample is equal, then mean is ${\frac {1}{n}}\sum x$ In this problem, P(X=x)=n for all x

hyuckdae
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In the simplest case a sample of size $n$ is a realisation of $(X_1,\ldots,X_n)$, where each $X_i$ is an independent copy of $X$. Then the sample average is $\bar{X}_n(\omega)=n^{-1}\sum_{i=1}^n X_i(\omega)$ (in your notation $X_i(\omega)=x_i$). The connection of this quantity to the expectation of $X$ is twofold (assuming that $\mathsf{E}|X|<\infty$): (1) $\mathsf{E}\bar{X}_n=\mathsf{E}X$, and (2) $\mathsf{P}(\bar{X}_n\to \mathsf{E}X)=1$.