Basic coin toss problem to get 1 head, but with a different method

[Master1022]

Homework Statement

What is the expected number of tosses until we get

Relevant Equations

Expected value

Hi,

I was watching a video where the following formula, for expected value, was presented for a discrete probability distribution which cannot take negative values:
$$E[X] = \sum_{j = 1}^{\infty} P(X \geq j)$$
and I never saw this formula before and am trying to develop an intuition for it. I am trying to solve a basic problem with this method just so I understand how it works.

Question:
What is the expected number of times needed to flip a fair coin until we get 1 head?

Attempt:
So the term $P(X \geq j)$ means that we didn't get a head in the first $(j - 1)$ throws. Thus, $P(X \geq j) = 1 - P(X \leq j - 1) = 1 - \begin{pmatrix} j - 1 // 1 \end{pmatrix} \cdot \left( \frac{1}{2} \right) ^{1} \cdot \left( \frac{1}{2} \right) ^{(j - 1) - 1} = \left( \frac{1}{2} \right) ^{j - 1}$

Thus our summation becomes:
$$E[X] = \sum_{j = 1}^{\infty} \left( \frac{1}{2} \right) ^{j - 1} = 2 \cdot \sum_{j = 1}^{\infty} \left( \frac{1}{2} \right) ^{j} = 2 \cdot \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 2$$

Is that a correct way to use this different formula? Is there any intuition behind this formula?

Thanks in advance.

 

[andrewkirk]

Yes, your application is correct. I can't follow your first line of formulas, which seems to have formatting problems, but your last line is correct, as is your conclusion.

For intuition, consider the following, where $p(x)$ is the probability that $X=x$.
The expected value of $X$ is
$$\sum_{x=1}^\infty x\cdot p(x)
=1\cdot p(1) + 2\cdot p(2) + 3\cdot p(3) + ...
= p(1) + p(2) + p(2) + p(3) + p(3) + p(3) + ...$$
We can arrange the terms in that sum in a triangle as follows:

p(1)
p(2) p(2)
p(3) p(3) p(3)
p(4) p(4) p(4) p(4)
.......
etc
The sum of items in the j-th row in the triangle equals the j-th term in the above sum.
And the sum of items in the j-th column is p(j) + p(j+1) + p(j+2) + ...
which equals $P(X\ge j)$.
So the formula you've been given corresponds to first adding each column and then adding the column sums together.
This contrasts to the usual formula, in which we first add each row, then add the row sums together.

In other words, we've just changed the order of summation in our infinite triangle of terms.
We are allowed to do this because all the terms are positive. We say the sum is "absolutely convergent". If some were negative, changing the order of summation could change the result.

 

[Master1022]

Thanks @andrewkirk !

Yes, your application is correct. I can't follow your first line of formulas, which seems to have formatting problems, but your last line is correct, as is your conclusion.

Apologies, it was supposed to be a combinatorics symbol:
$P(X \geq j) = 1 - P(X \leq j - 1) = 1 - \begin{pmatrix} j - 1 \\ 1 \end{pmatrix} \cdot \left( \frac{1}{2} \right) ^{1} \cdot \left( \frac{1}{2} \right) ^{(j - 1) - 1} = \left( \frac{1}{2} \right) ^{j - 1}$

I need to double check the algebra that follows after that line, but your response with the intuition was the main thing I was looking for.

For intuition, consider the following, where $p(x)$ is the probability that $X=x$.
The expected value of $X$ is
$$\sum_{x=1}^\infty x\cdot p(x)
=1\cdot p(1) + 2\cdot p(2) + 3\cdot p(3) + ...
= p(1) + p(2) + p(2) + p(3) + p(3) + p(3) + ...$$
We can arrange the terms in that sum in a triangle as follows:

p(1)
p(2) p(2)
p(3) p(3) p(3)
p(4) p(4) p(4) p(4)
.......
etc
The sum of items in the j-th row in the triangle equals the j-th term in the above sum.
And the sum of items in the j-th column is p(j) + p(j+1) + p(j+2) + ...
which equals $P(X\ge j)$.
So the formula you've been given corresponds to first adding each column and then adding the column sums together.
This contrasts to the usual formula, in which we first add each row, then add the row sums together.

In other words, we've just changed the order of summation in our infinite triangle of terms.
We are allowed to do this because all the terms are positive. We say the sum is "absolutely convergent". If some were negative, changing the order of summation could change the result.

That's actually quite interesting. Thank you very much for taking the time to write this out. I will have a think about it for a while, and respond if I have other questions. Otherwise, thanks!