Question 1

where is explained the sum of independent variables? E(X+Y) does not makes sense to me

Accepted Answer

Let
𝑋 = {𝑥₁, 𝑥₂, 𝑥₃}
𝑌 = {𝑦₁, 𝑦₂}

Thereby,
𝐸(𝑋) = (𝑥₁ + 𝑥₂ + 𝑥₃)∕3
𝐸(𝑌) = (𝑦₁ + 𝑦₂)∕2

Also,
𝑋 + 𝑌 = {𝑥₁ + 𝑦₁, 𝑥₂ + 𝑦₁, 𝑥₃ + 𝑦₁, 𝑥₁ + 𝑦₂, 𝑥₂ + 𝑦₂, 𝑥₃ + 𝑦₂}

This gives  us,
𝐸(𝑋 + 𝑌) = (𝑥₁ + 𝑦₁ + 𝑥₂ + 𝑦₁ + 𝑥₃ + 𝑦₁ + 𝑥₁ + 𝑦₂ + 𝑥₂ + 𝑦₂ + 𝑥₃ + 𝑦₂)∕6
= (2𝑥₁ + 2𝑥₂ + 2𝑥₃ + 3𝑦₁ + 3𝑦₂)∕6
= (𝑥₁ + 𝑥₂ + 𝑥₃)∕3 + (𝑦₁ + 𝑦₂)∕2
= 𝐸(𝑋) + 𝐸(𝑌)

This example can quite easily be generalized to where 𝑋 has 𝑚 elements and 𝑌 has 𝑛 elements.

Question 2

As the mean/expected value of a Bernoulli distribution is p and the mean/expected value of a binomial variable is np, is a binomial variable a multiple of a Bernoulli distribution?

Accepted Answer

Yes, the Bernoulli distribution is the simplest case of the binomial distribution where n = 1, meaning one trial.

Question 3

What is the expected value of a variable like:
"Flip a fair coin until you get tails. X = the number of heads you flipped."
I realize this wouldn't be a binomial variable, but it seemed pretty similar.

Note: P(H) = P(T) = 0.5

Accepted Answer

Never mind. I found it:
https://www.khanacademy.org/math/ap-statistics/random-variables-ap/geometric-random-variable/v/proof-of-expected-value-of-geometric-random-variable?modal=1

Question 4

Let X~Bin(n,p),find E(e^(tx) where t is a constant

Accepted Answer

Nice question!  The plan is to use the definition of expected value, use the formula for the binomial distribution, and set up to use the binomial theorem in algebra in the final step.

We have
E(e^(tx)) 
= sum over all possible k of P(X=k)e^(tk)
= sum k from 0 to n of p^k (1-p)^(n-k) (n choose k) e^(tk)
= sum k from 0 to n of (pe^t)^k (1-p)^(n-k) (n choose k)
= (pe^t + 1 - p)^n, from the binomial theorem in algebra.

Question 5

The thing I get caught up on is the Expected value of Y at 5:40. Could someone give me a link to the logic behind E(Y)=p? Specifically, when he talked about the probability of weighted outcomes?

Accepted Answer

It comes from the definition of expected value of a random variable Y: E[Y] = 0*p(Y=0) + 1*p(Y=1) + .. + n*p(Y=n). As defined by Sal in this example, the random variabla Y only takes the values Y=0 and Y=1. Moreover, we have that p(Y=1) = p and p(Y=0) = 1-p. This leads to E[Y] = p.

Question 6

Why is Y=0 or Y=1? 
If Y is not either 0 or 1, what kind of formula should we use?

Accepted Answer

In this case we want 𝑌 to represent whether a trial is a success or not.
So it needs to have two outcomes – one for "Success" and one for "Not Success".

The reason we choose the outcomes to be either 0 or 1 is because it allows us to easily count the number of successes after 𝑛 trials:
𝑌₁ + 𝑌₂ + 𝑌₃ + ... + 𝑌ₙ

– – –

The way we define a variable depends on what we want it to represent.

For example, if you and friend were competing in a game you might want to keep track of who has won more often after 𝑛 rounds.

Then we might want to define 𝑌 = −1 if you lose a round, 𝑌 = 0 if a round ends in a draw, and 𝑌 = 1 if you win a round.

If the sum 𝑌₁ + 𝑌₂ + 𝑌₃ + ... + 𝑌ₙ is negative you lost more rounds than you won,
If the sum is 0, then both of you won equally often.
And if the sum is positive you won more rounds than you lost.

Question 7

By assuming that E(nY) can be dissected into E(Y)+E(Y)+E(Y)+...+E(Y) n times, we have assumed that Y is independent with respect to itself, but clearly knowing what Y is would constrain what Y is, so are all random variables somehow independent with respect to themselves, and if so why?

Accepted Answer

The explanation given could be better.  X should really be written as the sum Y_1+Y_2+Y_3+...+Y_n, where Y_k is 1 if the kth trial is a success, and is 0 if the kth trial is a failure.  The variables Y_1,Y_2,Y_3,...Y_n are independent, and E(Y_k)=p for all k with 1<=k<=n.
So E(X) = E(Y_1+Y_2+Y_3+...+Y_n) = E(Y_1)+E(Y_2)+E(Y_3)+....+E(Y_n) = sum of n copies of p = np.

(By the way, the assumption of independence is actually not needed in this type of situation.  The expected value of the sum of *any* random variables is always the sum of their expected values.  However, the assumption of independence *would be needed* to make a similar statement about variance.)

Statistics and probability

Course: Statistics and probability > Unit 9

Expected value of a binomial variable

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