Variance of binomial distribution - 1 trial

[operationsres]

Homework Statement

For n trials, S_n can be seen as the sum of n independent single trials X_i, i = 1,2,...,n, with \mathbb{E}[X_i]=p and Var[X_i]=p(1-p).2. What I don't understand

I don't understand why Var[X_i]=p(1-p).

We know that: Var[X_i]=\mathbb{E}[(X_i - \mathbb{E}[X_i])^2] = \mathbb{E}[X_i^2 - 2X_i\mathbb{E}[X_i] + \mathbb{E}[X_i]^2] = \mathbb{E}[X_i^2] - \mathbb{E}[X_i]^2.

Taking \mathbb{E}[X_i]^2, we have \mathbb{E}[X_i]^2=p^2.

Taking \mathbb{E}[X_i^2], we have \mathbb{E}[X_i^2]=\mathbb{E}[\prod_{i=1}^2X_i] = \prod_{i=1}^2 \mathbb{E}[X_i] = \mathbb{E}[X_i]^2 = p^2.

So Var[X_i]= p^2 - p^2 = 0 \not= p(1-p), which contradicts what my lecture notes say.

 

[Ray Vickson]

Homework Statement

For n trials, S_n can be seen as the sum of n independent single trials X_i, i = 1,2,...,n, with \mathbb{E}[X_i]=p and Var[X_i]=p(1-p).


2. What I don't understand

I don't understand why Var[X_i]=p(1-p).

We know that: Var[X_i]=\mathbb{E}[(X_i - \mathbb{E}[X_i])^2] = \mathbb{E}[X_i^2 - 2X_i\mathbb{E}[X_i] + \mathbb{E}[X_i]^2] = \mathbb{E}[X_i^2] - \mathbb{E}[X_i]^2.

Taking \mathbb{E}[X_i]^2, we have \mathbb{E}[X_i]^2=p^2.

Taking \mathbb{E}[X_i^2], we have \mathbb{E}[X_i^2]=\mathbb{E}[\prod_{i=1}^2X_i] = \prod_{i=1}^2 \mathbb{E}[X_i] = \mathbb{E}[X_i]^2 = p^2.

So Var[X_i]= p^2 - p^2 = 0 \not= p(1-p), which contradicts what my lecture notes say.

No, E X_i^2 \neq p^2. In fact, E f(X_i) = \sum_{k} P\{X_i = k\} f(k) for any function f, so we get E X_i^2 = p.

RGV

 

[operationsres]

Enlightening post, thank you.

I guess my error was assuming that X_i was independent in:
\mathbb{E}[X_i^2]=\mathbb{E}[\prod_{i=1}^2X_i] = \prod_{i=1}^2 \mathbb{E}[X_i] = \mathbb{E}[X_i]^2 = p^2.?But they're the same event so that would be non-sensical...

 

[voko]

Use the definition of variance of a discrete variable.