b:head_first_statistics:using_discrete_probability_distributions
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b:head_first_statistics:using_discrete_probability_distributions [2023/10/01 19:06] – [Theorem 4: Var(c) = 0] hkimscil | b:head_first_statistics:using_discrete_probability_distributions [2024/10/01 21:44] (current) – [Fat Dan changed his prices] hkimscil | ||
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__Pool puzzle__ | __Pool puzzle__ | ||
+ | |||
+ | \begin{eqnarray*} | ||
+ | \text {original win} & = & \text {c(0, 5, 10, 15, 20)} \\ | ||
+ | \text {original cost} & = & \text {c(1)} \\ | ||
+ | \text {X} & = & \text {the amount of money you receive} \\ | ||
+ | \end{eqnarray*} | ||
\begin{eqnarray*} | \begin{eqnarray*} | ||
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E(X) = -.77 and E(Y) = -.85. What is 5 * E(X) + 3? | E(X) = -.77 and E(Y) = -.85. What is 5 * E(X) + 3? | ||
- | $ 5 * E(X) = -3.85 $ | + | \begin{eqnarray*} |
- | $ 5 * E(X) + 3 = -0.85 $ | + | E(X) & = & -.77 \\ |
- | $ E(Y) = 5 * E(X) + 3 $ | + | E(Y) & = & E(5X+3) \; |
+ | & = & E(5X) + 3 \\ | ||
+ | & = & 5 E(X) + 3 \\ | ||
+ | & = & -0.85 \\ | ||
+ | \\ | ||
+ | \\ | ||
+ | Var(Y) & = & Var(5X+3) \\ | ||
+ | & = & Var(5X) + Var(3) \\ | ||
+ | & = & Var(5X) \\ | ||
+ | & = & 5^2 Var(X) \\ | ||
+ | & = & 67.4275 | ||
+ | \end{eqnarray*} | ||
- | $ 5 * Var(X) = 13.4855 $ | ||
- | $ 5^2 * Var(X) = 67.4275 $ | ||
- | $ Var(Y) = 5^2 * Var(X) $ | ||
\begin{eqnarray*} | \begin{eqnarray*} | ||
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</ | </ | ||
- | 그런데 | + | 위는 각각 계산한 것. 아래는 Y = X - 0.5를 적용해서 계산한 것 |
\begin{eqnarray*} | \begin{eqnarray*} | ||
E(X-a) & = & E(X) - a \\ | E(X-a) & = & E(X) - a \\ | ||
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Variance는 기본적으로 아래와 같다. 이 때 $X=c$ 라고 (c=상수) 하면 | Variance는 기본적으로 아래와 같다. 이 때 $X=c$ 라고 (c=상수) 하면 | ||
\begin{align} | \begin{align} | ||
- | Var(X) & = E[X − E(X)^2] \text{ | + | Var(X) & = E[(X − E(X))^2] \text{ |
& = E[(c-c)^2] \nonumber | & = E[(c-c)^2] \nonumber | ||
& = 0 | & = 0 | ||
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\begin{align*} | \begin{align*} | ||
& E[(X + Y)^2] = E[X^2 + 2XY + Y^2] = E[X^2] + 2E[XY] + E[Y^2] \\ | & E[(X + Y)^2] = E[X^2 + 2XY + Y^2] = E[X^2] + 2E[XY] + E[Y^2] \\ | ||
- | - & [E(X + Y)]^2 = [E(X) + E(Y)]^2 = E(X)^2 + 2E(X)E(Y) + E(Y^2) \\ | + | - & [E(X + Y)]^2 = [E(X) + E(Y)]^2 = E(X)^2 + 2E(X)E(Y) + E(Y)^2 \\ |
\end{align*} | \end{align*} | ||
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Var[(X+Y)] = | Var[(X+Y)] = | ||
& E[X^2] & + & 2E[XY] & + & E[Y^2] \\ | & E[X^2] & + & 2E[XY] & + & E[Y^2] \\ | ||
- | - & E(X)^2 & - & 2E(X)E(Y) & - & E(Y^2) \\ | + | - & E(X)^2 & - & 2E(X)E(Y) & - & E(Y)^2 \\ |
& Var[X] & + & 2 E[XY]-2E(X)E(Y) & + & Var[Y] \\ | & Var[X] & + & 2 E[XY]-2E(X)E(Y) & + & Var[Y] \\ | ||
\end{align*} | \end{align*} | ||
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\end{align} | \end{align} | ||
====== e.gs in R ====== | ====== e.gs in R ====== | ||
- | |||
R에서 이를 살펴보면 | R에서 이를 살펴보면 | ||
< | < | ||
+ | # variance theorem 4-1, 4-2 | ||
+ | # http:// | ||
+ | # need a function, rnorm2 | ||
+ | rnorm2 <- function(n, | ||
+ | |||
m <- 0 | m <- 0 | ||
- | v <- 4 | + | v <- 1 |
n <- 10000 | n <- 10000 | ||
set.seed(1) | set.seed(1) | ||
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v.12 <- var(x1 + x2) | v.12 <- var(x1 + x2) | ||
v.12 | v.12 | ||
- | ## should be near 2*v | + | ###################################### |
+ | ## v.12 should be near var(x1)+var(x2) | ||
###################################### | ###################################### | ||
## 정확히 2*v가 아닌 이유는 x1, x2가 | ## 정확히 2*v가 아닌 이유는 x1, x2가 | ||
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# var(2*x1) = 2^2 var(X1) | # var(2*x1) = 2^2 var(X1) | ||
v.11 | v.11 | ||
- | |||
</ | </ |
b/head_first_statistics/using_discrete_probability_distributions.1696154794.txt.gz · Last modified: 2023/10/01 19:06 by hkimscil