2
$\begingroup$

I asked a question on this site about the same scenario:

The probability of a head occurring when a biased coin is tossed is $p$, where $p < 1$. Let the random variable $X$ represent the number of tosses up to and including the first toss on which a tail occurs.

So I found out that the probability is $p^{n-1} (1-p)$

I now need to find out the $E(X)$ is (which I think is the same as the mean).

I tried tabulating the values and adding them but got nowhere. How do I get the $E(X)$ value?

The answer is meant to be $\frac{1}{1-p}$.

$\endgroup$
  • $\begingroup$ Yes $E(X)$ is the expectation, or also called the mean, of the RV $X$ and is computed for discrete positive integer-valued random variables by the formula (which ought to be in your textbook, notes, on wikipedia etc) $E(X)=\sum_{n=1}^\infty n P(X=n)$. $\endgroup$ – Nap D. Lover Jun 3 '19 at 1:55
  • $\begingroup$ Additionally, if this specific sum is giving you trouble, here is a hint: review geometric series from calculus. $\endgroup$ – Nap D. Lover Jun 3 '19 at 1:57
  • $\begingroup$ Is it correct to use the a1 / 1 - r formula? I tried this by substituting 1-p as a and p as r, but I could not get the right answer. $\endgroup$ – Christopher U Jun 3 '19 at 2:23
  • $\begingroup$ the generous hint in the answer given below should be enough to conclude with some thought, what do you think? $\endgroup$ – Nap D. Lover Jun 3 '19 at 16:09
1
$\begingroup$

Hint: To compute $\sum_{n=1}^\infty np^{n-1}$, try to use the fact: $$(\sum_{n=1}^\infty x^n)’=\sum_{n=1}^\infty nx^{n-1},$$ for $x\in [0,1)$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

A bit more machinery than perhaps required. Using the law of total expectation we have that $$ \begin{align} EX&=E(X\mid X>1)P(X>1)+E(X\mid X=1)P(X=1)\\ &=(1+EX)p+1(1-p)\tag{0}\\ &=1+pEX \end{align} $$ where in (0) we used the fact that $P(X=1)=1-p$ (tails on the first try), $E(X\mid X=1)=E(1)=1$ and that $P(X>1)=p$ since the event $(X>1)$ corresponds to heads on the first toss. Finally $E(X\mid X>1)=1+EX$ since we failed to get a tail on the first toss and the process starts anew thereafter. Hence $$ EX=\frac{1}{1-p}. $$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

$$P(X = k) = p^{k-1}\times (1-p)$$

This is a Hypergeometric Distribution.

$$E(X) = \sum_{k=1}^{\infty} k \times P(X=k)$$ $$\implies E(X) = \sum_{k=1}^{\infty} k \times p^{k-1}\times (1-p) \quad (1)$$

$$p\times E(X) = \sum_{k=1}^{\infty} k \times p^{k}\times (1-p) \quad (2)$$

Now do $(1)-(2)$.

$$(1-p) E(X) = (1-p)\times \left[\sum_{k=0}^{\infty} p^k\right]$$

$$\implies E(X) =\frac{1}{1-p}$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.