Confusion on Lecture "Central Limit Theorem - (Discrete Random Variable)

[dapao]

Lecture: [Central Limit Theorem - Discrete Random Variable]

Hello everyone,

In example 1 (coin toss) of the Central Limit Theorem lecture, I have trouble understand how the variance is equal to np(1 - p). I was able to calculate the mean, which is equal to np. Can anyone help me understand how to derive the variance?

[Alireza_Saei]

Hi @dapao

For a binomial random variable X with parameters n and p , the variance is calculated as:

\text{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2

Mean → You already calculated \mathbb{E}[X] = np.

Second Moment → For a binomial random variable, \mathbb{E}[X^2] = n^2p^2 + np(1-p).

Variance → Substituting these values:

\text{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2 = (n^2p^2 + np(1-p)) - (np)^2

Simplify:

\text{Var}(X) = np(1-p)

Hope this helps! feel free to ask if you need further assistance.

[dapao]

Thank you @Alireza_Saei so much for the clarification!

[Alireza_Saei]

You’re welcome! Happy to help

[rmwkwok]

Hello, @dapao,

In case you might find them interesting to read, here are a few other proofs for np. I would recommend the first two.

Cheers,
Raymond