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Let $X$ be a geometrically distributed random variable with parameter P. Compute the density of $X^2$.

The density of the random variable $X$ is

$$f_X (x) = p(1-p)^x \text{ for } x=0,1, \ldots,$$

so I thought it would be $p(1-p)^{x^2}$ for $X^2$. However, the answer of the book simply says

$$P(X^2 =k^2), \quad f(x) = p(1-p)^x, x=0,1, \ldots,$$

which says essentially that there is no change. Why? I don't understand the reasoning.

The book is "Introduction to Probability Theory" by Hoel, Port and Stone.

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Hint: Consider the random variable $Z$ that is equal to, say, $3$ with probability $0.4$, and is equal to $7$ with probability $0.6$. Now, what can you say about the random variable $Z^2$? Can you apply the same idea to your problem?

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  • $\begingroup$ We have $P(Z=3)= 0.4$ and $P(Z=7) = 0.6$, therefore $P(Z=3) + P(Z=7) = 1$ and they are complementary. Considering $Z^2$ gives $P(Z^2 =9) = 0.4$ and $P(Z^2 = 49) = 0.6$? A new sample space of sorts but with probability unchanged? $\endgroup$ – Mark Fantini Mar 19 '14 at 3:07
  • $\begingroup$ @Fantini Yes. The same idea goes for your problem. $\endgroup$ – Lord Soth Mar 19 '14 at 4:21

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