數學問題啊，幫幫手

首先要明白甚麼是variance
variance 代表 [sum up每一個 (data – mean )^2 ] / total number of data
舉例來說: Data : 2, 4, 6, 8, 10
mean = 6
variance = [ (2-6)^2 + (4-6)^2 + ...+ (10-6)^2 ] / 5

而Var (a + x )代表每一個data 都加上 a後的新variance
data會變了 (原本的data + a)
mean 會變了 (原本的mean + a)
所以variance 就會變了: {sum up每一個 [(data + a) – (mean +a )]^2 } / total number of data
兩個 a 會消去所以variance 不變


而 Var (ax) 代表每一個data 都乘上 a後的新variance
data會變了 (原本的data * a)
mean 會變了 (原本的mean * a)
所以variance 就會變了: {sum up每一個 [(data * a) – (mean *a )]^2 } / total number of data
考慮 [(data * a) – (mean *a )]^2
= [ a ( data – mean)]^2
= a^2 * (data –mean)^2

所以新variance就會是舊variance * a^2

希望解答到你的問題

兩個答案都assume左biased的variance (i.e. 分母是n)，由他的另一問題可見，他用的是unbiased的定義 (i.e. 跟excel一樣，分母是n - 1)。

oliverng0226的答案放成unbiased版本都說得通，用algebraic方法的可能會比較煩了(當然還是可以做的)。

I am not 100% about the following proof...

==================================

The 2 formulas is valid when x is the randow variable and 'a' is a constant

Firstly, note that

E(a+x) = E(a) + E(x) = a + E(x)
E(ax) = aE(x)

and

Var(y) = E(y^2) - [E(y)]^2

================================

(i) Var (a+x)

= E((a+x)^2) - [E(a+x)]^2

= E(a^2 + x^2 + 2ax) - [a + E(x)]^2

= E(a^2) + E(x^2) + E(2ax) - [a^2 + E(x)^2 + 2aE(x)]

= a^2 + E(x^2) + 2aE(x) - a^2 - E(x)^2 - 2aE(x)

= E(x^2) - E(x)^2

= Var(x)

================================

(ii) Var(ax)

= E((ax)^2) - [E(ax)]^2

= a^2E(x^2) - [aE(x)]^2

= a^2E(x^2) - a^2[E(x)]^2

= a^2 [E(x^2) - E(x)^2]

= a^2 Var(x)