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I am missing something that might be trivial in deriving the mean of the geometric distribution function by using expected value identity

$$ \sum_x x \theta (1-\theta)^{x-1}. $$

Michael Hardy
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JuanSancen
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3 Answers3

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Letting $\alpha=1-\theta$, we have $$ \sum_{x=1}^\infty x \theta (1-\theta)^{x-1} = \theta\sum_{x=1}^\infty x \alpha^{x-1} = \theta \sum_{x=1}^\infty \frac{d}{d\alpha} \alpha^x = \theta\frac{d}{d\alpha}\sum_{x=1}^\infty \alpha^x. $$ Sum the geometric series. You'll get an expression that you can readily differentiate with respect to $\alpha$. Finally, put $1-\theta$ back in place of $\alpha$.

Michael Hardy
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Hint: for $|r|<1$

$$\sum_{k=1}^{\infty} k r^k = \frac{r}{(1-r)^2}$$

Ron Gordon
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\begin{align} E(X)&=\sum_xx\theta(1-\theta)^{x-1}\\&=\theta(1-(1-\theta))^{-2}=\frac{1}{\theta} \end{align}

Alternative: Use Probability generating function(PGF)

Here $P_X(t)=E(t^X)=(1-(1-\theta)t)^{-1}$ if $|t|<\frac{1}{1-\theta}$

So $E(X)=P'_X(t)|_{t=1}=\frac{1}{\theta}$

Alternative: Use Moment generating function(MGF)

$M_X(t)=E(e^{tX})=(1-(1-\theta)e^t)^{-1}$ if $|t|<\log\left[\frac{1}{1-\theta}\right]$

So $E(X)=M'_X(t)|_{t=0}=\frac{1}{\theta}$

Argha
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