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So, I was going through Pattern Recognition and Machine learning by Bishop, page - 18, and I came across probability densities where I read about the transformation of a random variable. I couldn't understand what it actually means? For example, let's take a die.

Let the outcome of a die roll be represented by the random variable $X$. So, the probability distribution is as follows: $P(X=1)=1/6, P(X=2)=1/6, P(X=3)=1/6, P(X=4)=1/6, P(X=5)=1/6, P(X=6)=1/6$

Now if I make a non-linear transformation of $X$ say, ${{X}^{2}}$ what does this actually mean? From the definition of probability distribution, I think, we can say that the following is the probability distribution of the random variable ${{X}^{2}}$. $P({{X}^{2}}=1)=1/2, P({{X}^{2}}=4)=1/2, P({{X}^{2}}=9)=0, P({{X}^{2}}=16)=0, P({{X}^{2}}=25)=0, P({{X}^{2}}=36)=0$ But how do I interpret ${{X}^{2}}$ w.r.t. a die roll? Also, why would I do such a thing?

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    $\begingroup$ How did you get those probabilities for $X^2$? $\endgroup$
    – Glen_b
    Dec 15 '18 at 22:56
  • $\begingroup$ have you ever gambled? I could offer a game where I pay you the square of what shows on the die - what would the faitr price to enter the game? $\endgroup$
    – seanv507
    Dec 15 '18 at 22:57
  • $\begingroup$ That's why I said, "I think". I am not sure about it. And that's exactly why I am confused about what the transformation it means. $\endgroup$
    – KAY_YAK
    Dec 15 '18 at 23:31
  • $\begingroup$ $X^2$ is just the random variable whose value is the square of $X$. For example, if you roll a 2, then $X = 2$ and $X^2 = 4$. If you roll a $3$, then $X=3$ and $X^2 = 9$. $\endgroup$
    – littleO
    Dec 16 '18 at 13:05
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Your calculations are incorrect.

For $X>0$, the transformation $Y=X^2$ is strictly monotonic (and so invertible).

When you transform discrete random variables with a one-to-one transformation, the probabilities are unchanged, it's only that the values taken by the new variable that are different to those taken by the old variable.

X    1    2    3    4    5    6
Y    1    4    9   16   25   36
p   1/6  1/6  1/6  1/6  1/6  1/6

p.m.f. of X, the outcome of a fair die and its square

For a continuous variate, probabilities are associated with intervals, and analogously, those probabilities are also unchanged -- it's just the endpoints of each interval that move when you use a one-to-one transformation.

Now if I make a non-linear transformation of $X$ say, $X^2$ what does this actually mean?

You're now considering a variable which takes the value of the square of the value on the face of the die.

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But how do I interpret ${{X}^{2}}$ w.r.t. a die roll? Also, why would I do such a thing?

You don't. This is just an abstract example. But it has sense in many cases. Say, instead of working with data on counts in hundreds of thousands of people, you work with logarithm of the counts, so now it is helpful to know how to work with transformed random variable. Another example: for your experiment you measured the temperature in Celsius degrees, but you need it in Fahrenheit degrees.

As about your calculation, it is wrong. Dice has six sides, $X \in \{1,2,3,4,5,6\}$. So $X^2=9$ is possible for $X=3$, so the probability is not zero.

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