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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).
in Data Science & Statistics by Diamond (40,719 points) | 24 views

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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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