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Consider a random variable $X$ which can take any positive integer value (i.e. $1,2,3, \ldots)$. Its pmf is
$f(x)=\frac{c}{4^{x}} .$
Find the value $c$, find its cdf, and calculate $P(X<4)$. Try to find its expected value (tricky).
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To find the value of $c$, we solve the summation
$1=\sum_{x=1}^{\infty} \frac{c}{4^{x}}=c \sum_{x=1}^{\infty}\left(\frac{1}{4}\right)^{x}$
The power series converges to $\frac{1}{1-1 / 4}-1=\frac{4}{3}-1=\frac{1}{3}$, so $c=3$.
$P(X<4)=P(X=1)+P(X=2)+P(X=3)=\frac{3}{4}+\frac{3}{16}+\frac{3}{64}=\frac{63}{64} \text {. }$
The expected value is
$E(X)=\sum_{x=1}^{\infty} \frac{3 x}{4^{x}}$
Recall that
$1+2 x+3 x^{2}+4 x^{3}+\cdots=\frac{d}{d x}\left(1+x+x^{2}+x^{3}+\cdots\right)=\frac{1}{(1-x)^{2}}$
if $-1<x<1$. In our case, $x=\frac{1}{4}$, and the exponents are the variables. We want to re-express our expected value so it is in this form.
$E(X)=\sum_{x=1}^{\infty} \frac{3 x}{4^{x}}=\frac{3}{4}\left(1+2\left(\frac{1}{4}\right)+3\left(\frac{1}{4}\right)^{2}+4\left(\frac{1}{4}\right)^{3}+\cdots\right)=\frac{3}{4}\left(\frac{1}{(1-1 / 4)^{2}}\right)$
This evaluates to $\frac{4}{3}$.
by Diamond (40,719 points)

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