- #1

Master1022

- 611

- 117

- 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:

[tex] E[X] = \sum_{j = 1}^{\infty} P(X \geq j) [/tex]

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.

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

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:

[tex] 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 [/tex]

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

Thanks in advance.

I was watching a video where the following formula, for expected value, was presented for a discrete probability distribution which cannot take negative values:

[tex] E[X] = \sum_{j = 1}^{\infty} P(X \geq j) [/tex]

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:

[tex] 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 [/tex]

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

Thanks in advance.