Solutions - 5120-03

(1) State the PMF of a geometric random variable:

$$P_X(k)={(1-p)}^{k-1}p$$

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(2) Use formula for expectation to find $E[X]$:

$$\begin{array}{cc}E[X]& =\sum _{k=1}^{\infty }k{(1-p)}^{k-1}p\\ & =p\sum _{k=1}^{\infty }k{(1-p)}^{k-1}\end{array}$$

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(3) Apply hint:

We have that $\frac{1}{1-x}={\sum }_{k=0}^{\infty }{x}^k$.

Differentiating both sides yields $\frac{1}{{(1-x)}^2}={\sum }_{k=1}^{\infty }k{x}^{k-1}$.

Note that here, $x=(1-p)$, so

$$E[X]=p\frac{1}{{(1-(1-p))}^2}=\frac{p}{p^2}=\frac{1}{p}$$

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(4) Find expression for $E[X^2]$:

Note that $\text{Var}[X]=E\left[X^2\right]-{\left(E[X]\right)}^2$.

Applying the hint, we have $E[X^2]=E[X(X-1)+X]$.

Using the linearity of expectation, we can write this as $E[X(X-1)]+E[X]$.

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(5) Find $E[X(X-1)]$ and $E[X^2]$:

First, note that the second derivative of $\frac{1}{1-x}$ is $\frac{2}{{(1-x)}^3}={\sum }_{k=2}^{\infty }k(k-1)x^{k-2}$.

$$E[X(X-1)]=\sum _{k=1}^{\infty }k(k-1){(1-p)}^{k-1}p=\frac{2-p}{p^2}-\frac{1}{p}$$

Thus, $E[X^2]=\frac{2-p}{p^2}-\frac{1}{p}+\frac{1}{p}=\frac{2-p}{p^2}$.

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(6) Find $\text{Var}[X]$:

$$\text{Var}[X]=\frac{2-p}{p^2}-\frac{1}{p^2}=\frac{1-p}{p^2}$$