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--- b:head_first_statistics:geometric_binomial_and_poisson_distributions [2025/10/06 11:57] – [Bernoulli Distribution] hkimscil
+++ b:head_first_statistics:geometric_binomial_and_poisson_distributions [2025/10/14 23:31] (current) – [e.g.,] hkimscil
@@ Line 73: / Line 73: @@
 ## rather than p * q^(r-1)
 dgeom(x = 0:n, prob = p)
-hist(dgeom(x = 0:n, prob = p))
+# hist(dgeom(x = 0:n, prob = p))
+barplot(dgeom(x=0:n, p))
 </code>
@@ Line 87: / Line 88: @@
 [29] 0.0003868563 0.0003094850
 >
-> hist(dgeom(x = 0:n, prob = p))
+> # hist(dgeom(x = 0:n, prob = p))
+> barplot(dgeom(x=0:n, p))
 </code>
-{{:b:head_first_statistics:pasted:20191030-023820.png}}
+<code> {{:b:head_first_statistics:pasted:20191030-023820.png}} </code>
+{{:b:head_first_statistics:pasted:20251013-080224.png}}
 r번 시도한 이후, 그 이후 어디서든지 간에 성공을 얻을 확률
 $$ P(X > r) = q^{r} $$
@@ Line 737: / Line 739: @@
 $Var(X) = \displaystyle \frac{q}{p^{2}}$
+<code>
+> p <- .4
+> q <- 1-p
+>
+> p*q^(2-1)
+[1] 0.24
+> dgeom(1, p)
+[1] 0.24
+>
+> 1-q^4
+[1] 0.8704
+> dgeom(0:3, p)
+[1] 0.4000 0.2400 0.1440 0.0864
+> sum(dgeom(0:3, p))
+[1] 0.8704
+> pgeom(3, p)
+[1] 0.8704
+>
+> q^4
+[1] 0.1296
+> 1-sum(dgeom(0:3, p))
+[1] 0.1296
+> 1-pgeom(3, p)
+[1] 0.1296
+> pgeom(3, p, lower.tail = F)
+[1] 0.1296
+>
+> 1/p
+[1] 2.5
+>
+> q/p^2
+[1] 3.75
+>
+</code>
@@ Line 787: / Line 822: @@
 \begin{eqnarray*}
 P(X = r) & = & _{n}C_{r} \cdot p^{r} \cdot q^{n-r} \;\;\; \text{Where,} \\
-_{n}C_{r} & = & \frac {n!}{r!(n-r)!}
+\displaystyle _{n}C_{r} & = & \displaystyle \dfrac {n!}{r!(n-r)!} \\
+\text{c.f.,  } \\
+\displaystyle _{n} P_{r} & = & \displaystyle \dfrac {n!} {(n-r)!} \\
 \end{eqnarray*}
+see [[:b:head_first_statistics:permutation_and_combination#what_if_horse_order_doesn_t_matter|Permutation chapter]]
 p = 각 시행에서 성공할 확률
@@ Line 877: / Line 917: @@
 c <- choose(n,r)
 ans1 <- c*(p^r)*(q^(n-r))
-ans1
+ans1    # or
+choose(n, r)*(p^r)*(q^(n-r))
+dbinom(r, n, p)
+# dbinom(2, 5, 1/4)
 </code>
 <code>
 > p <- .25
@@ Line 888: / Line 935: @@
 > ans <- c*(p^r)*(q^(n-r))
 > ans
+[1] 0.2636719
+>
+> choose(n, r)*(p^r)*(q^(n-r))
+[1] 0.2636719
+>
+> dbinom(r, n, p)
 [1] 0.2636719
 >
@@ Line 903: / Line 956: @@
 ans2 <- c*(p^r)*(q^(n-r))
 ans2
+choose(n, r)*(p^r)*(q^(n-r))
+dbinom(r, n, p)
 </code>
 <code>
@@ Line 914: / Line 972: @@
 > ans2
 [1] 0.08789062
+>
+> choose(n,r)*(p^r)*(q^(n-r))
+[1] 0.08789062
+>
+> dbinom(r, n, p)
+[1] 0.08789063
+>
 >
 </code>
-Ans 3.
+Ans 3. 중요
 <code>
 ans1 + ans2
+dbinom(2, 5, .25) + dbinom(3, 5, .25)
+dbinom(2:3, 5, .25)
+sum(dbinom(2:3, 5, .25))
+pbinom(3, 5, .25) - pbinom(1, 5, .25)
 </code>
-<code>> ans1 + ans2
+<code>
+> ans1 + ans2
 [1] 0.3515625
+> dbinom(2, 5, .25) + dbinom(3, 5, .25)
+[1] 0.3515625
+> dbinom(2:3, 5, .25)
+[1] 0.26367187 0.08789063
+> sum(dbinom(2:3, 5, .25))
+[1] 0.3515625
+> pbinom(3, 5, .25) - pbinom(1, 5, .25)
+[1] 0.3515625
+>
 </code>
@@ Line 979: / Line 1058: @@
 > </code>
-===== Another way to see E(X) and Var(X) =====
+Q. 한 문제를 맞힐 확률은 1/4 이다. 총 여섯 문제가 있다고 할 때, 0에서 5 문제를 맞힐 확률은? dbinom을 이용해서 구하시오.
+<code>
+p <- 1/4
+q <- 1-p
+n <- 6
+pbinom(5, n, p)
+- dbinom(6, n, p)
+sum(dbinom(0:5, n, p))
+</code>
+<code>
+> p <- 1/4
+> q <- 1-p
+> n <- 6
+> pbinom(5, n, p)
+[1] 0.9997559
+> 1 - dbinom(6, n, p)
+[1] 0.9997559
+> sum(dbinom(0:5, n, p))
+[1] 0.9997559
+>
+</code>
-==== Expectation and Variance value ====
+중요 . . . .
-\begin{eqnarray*}
+<code>
-E(X) & = & \sum_{x}xP(x) \\
+# http://commres.net/wiki/mean_and_variance_of_binomial_distribution
-& = & 0*p^{0}(1-p)^{1-0} + 1*p^{1}(1-p)^{1-1}  \\
+# ##################################################################
-& = & p \\
+#
-\\
+p <- 1/4
-Var(X) & = & E((X-\mu)^{2}) \\
+q <- 1 - p
-& = & \sum_{x}(x-\mu)^2P(x) \\
+n <- 5
-\end{eqnarray*}
+r <- 0
-그런데
+all.dens <- dbinom(0:n, n, p)
-\begin{eqnarray*}
+all.dens
-E((X-\mu)^{2}) & = & E(X^2) - (E(X))^2 \\
+sum(all.dens)
-\end{eqnarray*}
-위에서
+choose(5,0)*p^0*(q^(5-0))
-\begin{eqnarray*}
+choose(5,1)*p^1*(q^(5-1))
-E(X^{2}) & = & \sum x^2 p(x) \\
+choose(5,2)*p^2*(q^(5-2))
-& = & 0^2*p^0(1-p)^{1-0} + 1^2*p^1(1-p)^{1-1} \\
+choose(5,3)*p^3*(q^(5-3))
-& = & p
+choose(5,4)*p^4*(q^(5-4))
-\end{eqnarray*}
+choose(5,5)*p^5*(q^(5-5))
+all.dens
-zero squared probability of zero occurring
+choose(5,0)*p^0*(q^(5-0)) +
-one squared prob of one occurring
+  choose(5,1)*p^1*(q^(5-1)) +
+  choose(5,2)*p^2*(q^(5-2)) +
+  choose(5,3)*p^3*(q^(5-3)) +
+  choose(5,4)*p^4*(q^(5-4)) +
+  choose(5,5)*p^5*(q^(5-5))
+sum(all.dens)
+#
+(p+q)^n
+# note that n = whatever, (p+q)^n = 1
-또한 $E(X) = p $ 임을 알고 있음
+</code>
-\begin{eqnarray*}
-Var(X) & = & E((X-\mu)^{2}) \\
-& = & E(X^2) - (E(X))^2 \\
-& = & p - p^2 \\
-& = & p(1-p)
-\end{eqnarray*}
-위는 First Head Statistics 에서 $X \sim (1, 0.25)$ 에서 E(X)와 Var(X)를 구한 후 (각각, p와 pq), X가 n가지가 있다고 확장하여 np와 npq를 구한 것과 같다. 즉, 교재는 Bernoulli distribution을 이야기(설명)하지 않고, 활용하여 binomial distribution의 기대값과 분산값을 구해낸 것이다.
+<code>
+> # http://commres.net/wiki/mean_and_variance_of_binomial_distribution
-==== extension of Bernoulli Distribution ====
+> # ##################################################################
+> #
-$E(U_{i}) = p$  and $Var(U_{i}) = p(1-p)$ or $Var(U_{i}) = p \cdot q$
+> p <- 1/4
+> q <- 1 - p
-$$X = U_{1} + . . . . + U_{n}$$
+> n <- 5
-\begin{eqnarray*}
+> r <- 0
-E(X) & = & E(U_{1} + . . . + U_{n}) \\
+> all.dens <- dbinom(0:n, n, p)
-& = & E(U_{1}) + . . . + E(U_{n}) \\
+> all.dens
-& = & p + . . . + p \\
+[1] 0.2373046875 0.3955078125 0.2636718750 0.0878906250
-& = & np
+[5] 0.0146484375 0.0009765625
-\end{eqnarray*}
+> sum(all.dens)
+[1] 1
-\begin{eqnarray*}
+>
-Var(X) & = & Var(U_{1} + . . . + U_{n}) \\
+> choose(5,0)*p^0*(q^(5-0))
-& = & Var(U_{1}) + . . . + Var(U_{n}) \\
+[1] 0.2373047
-& = & p(1-p) + . . . + p(1-p) \\
+> choose(5,1)*p^1*(q^(5-1))
-& = & np(1-p) \\
+[1] 0.3955078
-& = & npq
+> choose(5,2)*p^2*(q^(5-2))
-\end{eqnarray*}
+[1] 0.2636719
+> choose(5,3)*p^3*(q^(5-3))
+[1] 0.08789062
-===== From a scratch (Proof of Binomial Expected Value) =====
+> choose(5,4)*p^4*(q^(5-4))
+[1] 0.01464844
+> choose(5,5)*p^5*(q^(5-5))
+[1] 0.0009765625
+> all.dens
+[1] 0.2373046875 0.3955078125 0.2636718750 0.0878906250
+[5] 0.0146484375 0.0009765625
+>
+> choose(5,0)*p^0*(q^(5-0)) +
++   choose(5,1)*p^1*(q^(5-1)) +
++   choose(5,2)*p^2*(q^(5-2)) +
++   choose(5,3)*p^3*(q^(5-3)) +
++   choose(5,4)*p^4*(q^(5-4)) +
++   choose(5,5)*p^5*(q^(5-5))
+[1] 1
+> sum(all.dens)
+[1] 1
+> #
+> (p+q)^n
+[1] 1
+> # note that n = whatever, (p+q)^n = 1
+>
+</code>
+===== Proof of Binomial Expected Value and Variance =====
 [[:Mean and Variance of Binomial Distribution|이항분포에서의 기댓값과 분산에 대한 수학적 증명]], Mathematical proof of Binomial Distribution Expected value and Variance
 ====== Poisson Distribution ======