Why Cant We Directly Find The PDF Of The Transformation Of Random?

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Why can't we directly find the PDF of the transformation of random variables, say g(X) from the random variable X. Why do we have to first convert PDF into CDF and then have to differentiate to get to the PDF of the transformed random variable?

The distribution of X is the uniform(0, 1) distribution, and the distribution of Y is the uniform(0, 2) distribution. You don’t say so, but presumably you want to add the assumption that X and Y are independent, otherwise there is little you can do. Clearly, Z = max(X, Y) lies between 0 and 2. It suffices to compute Prob(Z < z) for z between 0 and 2. (No need to distinguish “less than” and “less than or equals to” so I’ll just write <, makes typing faster). It would be smart to distinguish between values of z in (0, 1) and values of z in (1, 2). Z is smaller than z if and only if X and Y are both smaller than z. For z in (0, 1) the probability of that event is z times z/2. For z in (1, 2) it’s 1 times z/2. So the cdf takes the form z^2/2 then z/2 on those two intervals and the density is z on (0, 1) and 1/2 on (1, 2); of course it is zero outside of (0, 2).

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Convert PDF: All You Need to Know

Now we want all the other distributions. If the density is not equal to 1 on each interval then there is no distribution on the right. A distribution with (0, 1) = 0 and z = 1 on (0, 1) is clearly Z = max(X, Y), and Z = max(X, Y) > 1 — (1/2)Z/2 > 1/3 Then there is no distribution on the right. (0, 1) = 0 and z = 1 on (0, 1) is Clearly no distributions. Similarly, for (1, 2) = 0 and z = 1 on (0, 1) is also not a distribution. So for the (0 to 1) interval there are only two. Then the CDF is just the above two cases in reverse. (0, 1) = 0 and z = X on (0, 1) is clearly the CDF. Similarly, for (1, 2) = 0 and z = 1 on (0, 1) is clearly also the CDF. Clearly (0, 1) = 0.

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