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I'm given the formula:

$(1-p)^{x-1}p, x = 1, 2, ..., \infty$

and I'm asked to find the mean and variance.

I know the mean is represented by $\sum_{i=1}^n p_ix_i$ and the variance by $\sum_{i=1}^n p_i(x_i-\mu)^2$, but I'm not really sure how to get from those formulas to a generalized answer for the mean and variance. I've looked around online, but nearly everything seems to involve the use of a finite set of numbers. Those examples make sense, but I'm not sure how to translate that understanding into the more general solution asked for above. An explanation or a link to reading on the topic would help greatly!

Ryan
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2 Answers2

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Calculating the expected value

$$E(X)=p\sum_{x=1}^{\infty} x\cdot (1-p)^{x-1} $$

  • Indexshift: $x=k+1 \Rightarrow k=x-1$

$$E(X)=p\sum_{k=0}^{\infty} (k+1)\cdot (1-p)^{k}=p\sum_{k=0}^{\infty} k\cdot (1-p)^{k}+p\sum_{k=0}^{\infty} (1-p)^{k}$$

  • The first summand of $p\sum_{k=0}^{\infty} k\cdot (1-p)^{k}$ is 0.

Thus $$p\sum_{k=0}^{\infty} k\cdot (1-p)^{k}=p\sum_{k=1}^{\infty} k\cdot (1-p)^{k}$$

$$E(X)=p\sum_{k=1}^{\infty} k\cdot (1-p)^{k}+p\sum_{k=0}^{\infty} (1-p)^{k}$$

  • Multiplying out (1-p) at the first sigma sign.

$$E(X)=(1-p) \underbrace{p\sum_{k=1}^{\infty} k\cdot (1-p)^{k-1}}_{E(x)}+\underbrace{p\sum_{k=0}^{\infty} (1-p)^{k}}_{=1}$$

Thus $E(x)=(1-p)\cdot E(x)+1 \Rightarrow E(x)=\frac{1}{p}$

callculus42
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I'm given the formula: $$(1−p)^{x−1} p \quad,x=1,2,...,∞ $$

That's not a formula. A formula would be: $$P_X(x)=(1-p)^{x-1}p\quad:x\in\{1,2\ldots,\infty\}$$

and I'm asked to find the mean and variance.

I know the mean is represented by $\sum_{i=1}^n p_i x_i$ and the variance by $\sum^n_{i=1} p_i (x_i −μ)^2$.

In this case $$\begin{align} \mathsf E(X) & = \sum_{x=1}^\infty x\mathsf P_X(x) \\[2ex] \mathsf {Var}(X) & =\sum_{x=1}^\infty (x-\mathsf E(X))^2\mathsf P_X(x) \\ & = \sum_{x=1}^\infty x^2\mathsf P_X(x) - (\mathsf E(x))^2 \end{align}$$ Now $$\begin{align} \mathsf E(x) & = \sum_{x=1}^\infty x\mathsf P_X(x) \\ & = \sum_{x=1}^\infty xp(1-p)^{x-1} \\ & = p(1+\frac 2{(1-p)}+\frac 3{(1-p)^2} + \frac 4{(1-p)^3} + \cdots) \end{align}$$

Which is based on an order -1 polylogarithm (an extension of geometric series). $\sum_{k=1}^\infty k r^k = r/(1-r)^2, |r|<1$

Can you complete?

Graham Kemp
  • 129,094
  • This is not a geometric series. If you had $1 +\dfrac1{1-p}+\dfrac1{(1-p)^2}+\dfrac1{(1-p)^3}+\cdots$, that would be a geometric series. ${}\qquad{}$ – Michael Hardy Oct 29 '14 at 01:56