-
+
+{% include proof/start.html id="proof-mean" -%}
The trick is to treat $$p$$ and $$q$$ as independent until the last moment:
$$\begin{aligned}
@@ -62,8 +59,8 @@ $$\begin{aligned}
\end{aligned}$$
Inserting $$q = 1 - p$$ then gives the desired result.
-
-
+{% include proof/end.html id="proof-mean" %}
+
Meanwhile, we find the following variance $$\sigma^2$$,
with $$\sigma$$ being the standard deviation:
@@ -74,12 +71,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
-We use the same trick to calculate $$\overline{n^2}$$
+
+{% include proof/start.html id="proof-var" -%}
(the mean squared number of successes):
$$\begin{aligned}
@@ -106,8 +99,8 @@ $$\begin{aligned}
\end{aligned}$$
By inserting $$q = 1 - p$$, we arrive at the desired expression.
-
-
+{% include proof/end.html id="proof-var" %}
+
As $$N \to \infty$$, the binomial distribution
turns into the continuous normal distribution,
@@ -119,11 +112,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-normal" -%}
We take the Taylor expansion of $$\ln\!\big(P_N(n)\big)$$
around the mean $$\mu = Np$$:
@@ -211,8 +201,7 @@ $$\begin{aligned}
\end{aligned}$$
Taking $$\exp$$ of this expression then yields a normalized Gaussian distribution.
-
-
+{% include proof/end.html id="proof-normal" %}
## References
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