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--- b:head_first_statistics:geometric_binomial_and_poisson_distributions [2025/10/06 20:53] – [Expectation and Variance of] hkimscil
+++ b:head_first_statistics:geometric_binomial_and_poisson_distributions [2025/10/06 23:43] (current) – [e.g.,] hkimscil
@@ Line 797: / Line 797: @@
 ===== Expectation and Variance of =====
+Toss a fair coin once. What is the distribution of the number of heads?
+  * A single trial
+  * The trial can be one of two possible outcomes -- success and failure
+  * P(success) = p
+  * P(failure) = 1-p
+X = 0, 1 (failure and success)
+$P(X=x) = p^{x}(1-p)^{1-x}$ or
+$P(x) = p^{x}(1-p)^{1-x}$
+참고.
+| x     | 0          | 1  |
+| p(x)  | q = (1-p)  | p  |
+When x = 0 (failure), $P(X = 0) = p^{0}(1-p)^{1-0} = (1-p)$ = Probability of failure
+When x = 1 (success), $P(X = 1) = p^{1}(1-p)^{0} = p $ = Probability of success
-{{:b:head_first_statistics:pasted:20191104-013651.png}}
 This is called Bernoulli distribution.
@@ Line 862: / Line 877: @@
 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)
 </code>
 <code>
 > p <- .25
@@ Line 873: / Line 894: @@
 > 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
 >
 >
 </code>
 Ans 2.
@@ Line 888: / Line 920: @@
 ans2 <- c*(p^r)*(q^(n-r))
 ans2
+choose(n, r)*(p^r)*(q^(n-r))
+dbinom(r, n, p)
 </code>
 <code>
@@ Line 899: / Line 936: @@
 > 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 964: / Line 1022: @@
 > </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.
-==== Bernoulli Distribution ====
+<code>
-Toss a fair coin once. What is the distribution of the number of heads?
+p <- 1/4
-  * A single trial
+q <- 1-p
-  * The trial can be one of two possible outcomes -- success and failure
+n <- 6
-  * P(success) = p
+pbinom(5, n, p)
-  * P(failure) = 1-p
-X = 0, 1 (failure and success)
+ - dbinom(6, n, p)
-$P(X=x) = p^{x}(1-p)^{1-x}$ or
+</code>
-$P(x) = p^{x}(1-p)^{1-x}$
+<code>
+> p <- 1/4
+> q <- 1-p
+> n <- 6
+> pbinom(5, n, p)
+[1] 0.9997559
+> 1 - dbinom(6, n, p)
+[1] 0.9997559
-참고.
+</code>
-| x  | 0  | 1  |
-| p(x)  | q = (1-p)  | p  |
-When x = 0 (failure), $P(X = 0) = p^{0}(1-p)^{1-0} = (1-p)$ = Probability of failure
+중요 . . . .
-When x = 1 (success), $P(X = 1) = p^{1}(1-p)^{0} = p $ = Probability of success
+<code>
+# http://commres.net/wiki/mean_and_variance_of_binomial_distribution
+# ##################################################################
+#
+p <- 1/4
+q <- 1 - p
+n <- 5
+r <- 0
+all.dens <- dbinom(0:n, n, p)
+all.dens
+sum(all.dens)
-<WRAP box>
+choose(5,0)*p^0*(q^(5-0))
-Bernoulli distribution expands to binomial distribution, geometric distribution, etc.
+choose(5,1)*p^1*(q^(5-1))
-Binomial distribution = The distribution of number of success in n independent Bernoulli trials.
+choose(5,2)*p^2*(q^(5-2))
-Geometric distribution = The distribution of number of trials to get the first success in independent Bernoulli trials.
+choose(5,3)*p^3*(q^(5-3))
-</WRAP>
+choose(5,4)*p^4*(q^(5-4))
+choose(5,5)*p^5*(q^(5-5))
+all.dens
-$P(X=x) = p^{x}(1-p)^{1-x}$ or
+choose(5,0)*p^0*(q^(5-0)) +
-$P(x) = p^{x}(1-p)^{1-x}$
+  choose(5,1)*p^1*(q^(5-1)) +
-X takes, x = 0, 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
-==== Expectation and Variance value ====
+</code>
-\begin{eqnarray*}
-E(X) & = & \sum_{x}xP(x) \\
-& = & 0*p^{0}(1-p)^{1-0} + 1*p^{1}(1-p)^{1-1}  \\
-& = & p \\
-\\
-Var(X) & = & E((X-\mu)^{2}) \\
-& = & \sum_{x}(x-\mu)^2P(x) \\
-\end{eqnarray*}
-그런데
-\begin{eqnarray*}
-E((X-\mu)^{2}) & = & E(X^2) - (E(X))^2 \\
-\end{eqnarray*}
-위에서
+<code>
-\begin{eqnarray*}
+> # http://commres.net/wiki/mean_and_variance_of_binomial_distribution
-E(X^{2}) & = & \sum x^2 p(x) \\
+> # ##################################################################
-& = & 0^2*p^0(1-p)^{1-0} + 1^2*p^1(1-p)^{1-1} \\
+> #
-& = & p
+> p <- 1/4
-\end{eqnarray*}
+> q <- 1 - p
+> n <- 5
-zero squared probability of zero occurring
+> r <- 0
-one squared prob of one occurring
+> all.dens <- dbinom(0:n, n, p)
+> all.dens
-또한 $E(X) = p $ 임을 알고 있음
+[1] 0.2373046875 0.3955078125 0.2636718750 0.0878906250
-\begin{eqnarray*}
+[5] 0.0146484375 0.0009765625
-Var(X) & = & E((X-\mu)^{2}) \\
+> sum(all.dens)
-& = & E(X^2) - (E(X))^2 \\
+[1] 1
-& = & p - p^2 \\
+>
-& = & p(1-p)
+> choose(5,0)*p^0*(q^(5-0))
-\end{eqnarray*}
+[1] 0.2373047
+> choose(5,1)*p^1*(q^(5-1))
-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.
+[1] 0.3955078
+> choose(5,2)*p^2*(q^(5-2))
-==== extension of Bernoulli Distribution ====
+[1] 0.2636719
+> choose(5,3)*p^3*(q^(5-3))
-$E(U_{i}) = p$  and $Var(U_{i}) = p(1-p)$ or $Var(U_{i}) = p \cdot q$
+[1] 0.08789062
+> choose(5,4)*p^4*(q^(5-4))
-$$X = U_{1} + . . . . + U_{n}$$
+[1] 0.01464844
-\begin{eqnarray*}
+> choose(5,5)*p^5*(q^(5-5))
-E(X) & = & E(U_{1} + . . . + U_{n}) \\
+[1] 0.0009765625
-& = & 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*}
+>
+> choose(5,0)*p^0*(q^(5-0)) +
-\begin{eqnarray*}
++   choose(5,1)*p^1*(q^(5-1)) +
-Var(X) & = & Var(U_{1} + . . . + U_{n}) \\
++   choose(5,2)*p^2*(q^(5-2)) +
-& = & Var(U_{1}) + . . . + Var(U_{n}) \\
++   choose(5,3)*p^3*(q^(5-3)) +
-& = & p(1-p) + . . . + p(1-p) \\
++   choose(5,4)*p^4*(q^(5-4)) +
-& = & np(1-p) \\
++   choose(5,5)*p^5*(q^(5-5))
-& = & npq
+[1] 1
-\end{eqnarray*}
+> sum(all.dens)
+[1] 1
+> #
-===== From a scratch (Proof of Binomial Expected Value) =====
+> (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 ======