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--- b:head_first_statistics:geometric_binomial_and_poisson_distributions [2023/10/17 18:05] – [What does the Poisson distribution look like?] hkimscil
+++ b:head_first_statistics:geometric_binomial_and_poisson_distributions [2023/10/19 18:55] – [Geometric Binomial and Poisson Distributions] hkimscil
@@ Line 1: / Line 1: @@
 ====== Geometric Binomial and Poisson Distributions ======
+정리
+\begin{align*}
+\text{Geometric Distribution:  } \;\;\; \text{X} & \thicksim Geo(p) \\
+p(X = k) & = q^{k-1} \cdot p \\
+E\left[ X \right] & = \frac{1}{p} \\
+V\left[ X \right] & = \frac{q}{p^2} \\
+\\
+\text{Binomial Distribution:  } \;\;\; \text{X} & \thicksim B(n, p) \\
+p(X = r) & = \binom{n}{r} \cdot p^{r} \cdot q^{n-r} \\
+E\left[ X \right] & = {n}{p} \\
+V\left[ X \right] & = {n}{p}{q} \\
+\\
+\text{Poisson Distribution:  } \;\;\; \text{X} & \thicksim P( \lambda ) \\
+P(X=r) & = e^{- \lambda} \dfrac{\lambda^{r}} {r!} \\
+E\left[ X \right] & = \lambda \\
+V\left[ X \right] & = \lambda \\
+\end{align*}
 ===== Geometric Distributions =====
@@ Line 924: / Line 942: @@
 ==== From a scratch (Proof of Binomial Expected Value) ====
-see [[:The Binomial Theorem]]
+[[:Mean and Variance of Binomial Distribution|이항분포에서의 기댓값과 분산에 대한 수학적 증명]], Mathematical proof of Binomial Distribution Expected value and Variance
-\begin{eqnarray*}
-\text{The binomial theorem} &  & \\
-(a + b)^{m} & = & \sum^{m}_{y=0}{{m}\choose{y}} a^{y} b^{m-y} \\
-\end{eqnarray*}
-위의 식이 복잡해 보이지만 m = 3 일때의 이항정리식을 말한다
-\begin{align*}
-\sum^{m}_{y=0}{{m}\choose{y}} a^{y} b^{m-y} \text{, m = 3} \\
-\end{align*}
-\begin{align*}
-\sum^{3}_{y=0}{{3}\choose{y}} a^{y} b^{3-y}
-& = {{3}\choose{0}} a^{0} b^{3-0}
-+ {{3}\choose{1}} a^{1} b^{3-1}
-+ {{3}\choose{2}} a^{2} b^{3-2}
-+ {{3}\choose{3}} a^{y} b^{3-3}  \\
-& = 1*a^0*b^3
-+ 3*a^1*b^2
-+ 3*a^2*b^1
-+ 1*a^3*b^0  \\
-& = a^3
-+ 3 a^2 b^1
-+ 3 a^1 b^2
-+ b^3  \\
-\end{align*}
-==== For Mean ====
-\begin{eqnarray*}
-E(X) & = & \sum_{x}x p(x) \\
-& = & \sum_{x=0}^{n} x {{n}\choose {x}} p^x(1-p)^{n-x}  \\
-& = & \sum_{x=0}^{n} x \frac{n!}{x!(n-x)!} p^x(1-p)^{n-x}  \\
-\text{note that   } x! = x(x-1)! \\
-& = & \sum_{x=1}^{n} \frac{n!}{(x-1)!(n-x)!} p^x(1-p)^{n-x}  \\
-\text{cause we know that E(x) = np,} \\
-\text{we extract np outside from summation} \\
-\text{note that  } p^x = p * p^{x-1} \\
-\text{and  } n! = n * (n-1)! \\
-& = & \sum_{x=1}^{n} \frac{\underline{n}*(n-1)!}{(x-1)!(n-x)!} (\underline{p}*p^{x-1})(1-p)^{n-x}  \\
-\text{we take out the underlined part} \\
-\text{(that is, np) out of the sigma part } \\
-& = & np \sum_{x=1}^{n} \frac{(n-1)!}{(x-1)!(n-x)!} p^{x-1}(1-p)^{n-x}  \\
-& = & np \underline{ \sum_{x=1}^{n} {\frac{(n-1)!}{(x-1)!(n-x)!} p^{x-1}(1-p)^{n-x}}}  \\
-\text{we want to check the underlined} \\
-\text{part is equal to one so that np is left out} \\
-n-x = (n-1)-(x-1) \\
-& = & np \sum_{x=1}^{n} \frac{(n-1)!}{(x-1)!((n-1)-(x-1))!} p^{x-1}(1-p)^{(n-1)-(x-1)}  \\
-m = n - 1 \\
-y = x - 1 \\
-& = & np \sum_{y=0}^{m} \frac{(m)!}{(y)!(m-y))!} p^{y}(1-p)^{(m-y)}  \\
-& = & np \sum_{y=0}^{m} \frac{(m)!}{(y)!(m-y))!} p^{y}(1-p)^{(m-y)}  \\
-& = & np \sum_{y=0}^{m} {{m}\choose {y}} p^{y}(1-p)^{(m-y)}  \\
-\text{Recall that} \\
-\sum^{m}_{y=0}{{m}\choose{y}} a^{y} b^{m-y} = (a + b)^{m} \\
-& = & np (p + (1-p))^m  \\
-& = & np (1)^m  \\
-& = & np  \\
-\end{eqnarray*}
-==== For variance ====
-\begin{align*}
-Var(X) & = E[(X-\mu)^2] \\
-& = \sum_{x}(x-\mu)^2p(x) \\
-\text{We also know that: } \\
-E[(X-\mu)^2] & = E(X^2) - [E(X)]^2 \\
-\end{align*}
-\begin{align*}
-E(X^2) & = \sum_{x=0}^{n} x^2 {{n}\choose{x}} p^x (1-p)^{n-x} \\
-& = \sum_{x=0}^{n} x^2 \frac{n!}{x!(n-x)!} p^x (1-p)^{n-x} \\
-\end{align*}
-그런데 $E(X^2)$ 대신 $E[X(X-1)]$을 생각해보면
-\begin{align*}
-E[X(X-1)] & = \sum_{x=0}^{n} x(x-1) \frac{n!}{x!(n-x)!} p^x (1-p)^{n-x} \\
-& = \sum_{x=2}^{n} \frac{n!}{(x-2)!(n-x)!} p^x (1-p)^{n-x} \\
-& = n(n-1)p^2 \sum_{x=2}^{n} \frac{(n-2)!}{(x-2)!(n-x)!} p^{x-2} (1-p)^{n-x} \\
-\text{cause} \\
-n - x = (n - 2)-(x - 2) \\
-& = n(n-1)p^2 \sum_{x=2}^{n} \frac{(n-2)!}{(x-2)!((n-2)-(x-2))!} p^{x-2} (1-p)^{(n-2)-(x-2)} \\
-m = n - 2 \\
-y = x - 2 \\
-& = n(n-1)p^2 \sum_{y=0}^{m} \frac{m!}{y!((m-y))!} p^{y} (1-p)^{(m-y)} \\
-& = n(n-1)p^2 \underline {\sum_{y=0}^{m} \frac{m!}{y!((m-y))!} p^{y} (1-p)^{m-y} } \\
-& = n(n-1)p^2 \underline {\sum_{y=0}^{m} {{m}\choose{y}} p^{y} (1-p)^{m-y} } \\
-\text {we know that the underline part is} \\
-(p+(1-p))^m = 1^m \\
-& = n(n-1)p^2 (p + (1-p))^m \\
-& = n(n-1)p^2 \\
-\end{align*}
-. . . .
-\begin{align*}
-E[X(X - 1)] & = n(n-1)p^2 \\
-E[X^2 - X] & = n(n-1)p^2 \\
-E[X^2]- E[X] & = n(n-1)p^2 \\
-E[X^2]- np & = n(n-1)p^2 \\
-E[X^2]& = n(n-1)p^2 + np \\
-\\
-\end{align*}
-\begin{align*}
-Var(X) & = E[X^2] - [E(X)]^2  \\
-& = \left[n(n-1)p^2 + np \right] - \left[np \right]^2  \\
-& = np[(n-1)p + 1 - np]  \\
-& = np[np - p + 1 - np]  \\
-& = np(1 - p)  \\
-& = npq  \\
-\\
-\end{align*}
 ====== Poisson Distribution ======
 $$X \sim Po(\lambda)$$
@@ Line 1121: / Line 1024: @@
 \end{eqnarray*}
-에서 1 이란 이야기는 아래 그림의 그래프가 전체가 1이 됨을 의미함. 즉 위에서는 1부터 60까지 갔지만, 1부터 무한대로 하면 완전한 분포곡선이 되는데 이것이 1이라는 뜻 (가령 dpois(x=1:10000000000, lambda=30)과 같은 케이스).
+에서 1 이란 이야기는 아래 그림의 그래프가 전체가 1이 됨을 의미함. 즉 위에서는 1부터 60까지 갔지만, 1부터 무한대로 하면 완전한 분포곡선이 되는데 이것이 1이라는 뜻 (가령 dpois(x=1:1000, lambda=30)과 같은 케이스).
@@ Line 1196: / Line 1099: @@
 > dpois(x=3, lambda=3.4)
 [1] 0.2186172
+</code>
+마찬가지로 적어도 3번까지 고장나는 경우는 0, 1, 2, 3을 포함하므로
+<code>
+> sum(dpois(c(0:3), lambda=3.4))
+[1] 0.5583571
+>
 </code>
@@ Line 1282: / Line 1192: @@
 > a*b*c
 [1] 0.03268244
+>
+</code>
+위가 답이긴 하지만 limited calculator 로는
+x ~ b (n, p)이고
+b(100, 0.1)이므로
+n*p = 10 = lambda
+따라서
+<code>
+> dpois(x=15, lambda=10)
+[1] 0.03471807
 >
 </code>