Skip to main content
deleted 6 characters in body
Source Link
juod
  • 2.3k
  • 14
  • 20

Note that the values of $X \sim Binomial(n,p)$ correspond to number of "positive" trials, not probability. As $n$ grows, the values of $\hat{p} = X/n$ converge to the true $p$, hence the probability as a long-run frequency definition. Values of $\bar{x} = n\hat{p}$$X = n\hat{p}$ clearly don't converge.

Same distinction might help understanding the variance: variance of the proportion of positive trials decreases, $Var(X/n) = np(1-p) / n^2 = p(1-p)/n$, so the proportion estimates get more precise. Variance of the actual number of positive outcomes increases, however.

Note that the values of $X \sim Binomial(n,p)$ correspond to number of "positive" trials, not probability. As $n$ grows, the values of $\hat{p} = X/n$ converge to the true $p$, hence the probability as a long-run frequency definition. Values of $\bar{x} = n\hat{p}$ clearly don't converge.

Same distinction might help understanding the variance: variance of the proportion of positive trials decreases, $Var(X/n) = np(1-p) / n^2 = p(1-p)/n$, so the proportion estimates get more precise. Variance of the actual number of positive outcomes increases, however.

Note that the values of $X \sim Binomial(n,p)$ correspond to number of "positive" trials, not probability. As $n$ grows, the values of $\hat{p} = X/n$ converge to the true $p$, hence the probability as a long-run frequency definition. Values of $X = n\hat{p}$ clearly don't converge.

Same distinction might help understanding the variance: variance of the proportion of positive trials decreases, $Var(X/n) = np(1-p) / n^2 = p(1-p)/n$, so the proportion estimates get more precise. Variance of the actual number of positive outcomes increases, however.

Source Link
juod
  • 2.3k
  • 14
  • 20

Note that the values of $X \sim Binomial(n,p)$ correspond to number of "positive" trials, not probability. As $n$ grows, the values of $\hat{p} = X/n$ converge to the true $p$, hence the probability as a long-run frequency definition. Values of $\bar{x} = n\hat{p}$ clearly don't converge.

Same distinction might help understanding the variance: variance of the proportion of positive trials decreases, $Var(X/n) = np(1-p) / n^2 = p(1-p)/n$, so the proportion estimates get more precise. Variance of the actual number of positive outcomes increases, however.