Find the probability density function for $Y = X^2 + 1$

Question

The question is, let $X$ be a random variable, on $[0, 1]$, with probability density function $p(x) = 2-2x$.

Let $Y$ be a random variable on $[1, 2]$, such that $Y = X^2 + 1$. Find the pdf for $Y$.

I was wondering how the statement "$Y$ on $[1, 2]$" has effects on the solution. The general approach would be constructing corresponding CDF and then differentiate it. I wasn't too sure if the statement has any effects on the solution.

Thank you in advance!

Answer 1

Easy use the fundamental transformation theorem

$$f_Y(y)=f_X(g^{-1}(y))\Bigg|\frac{d}{dy}g^{-1}(y)\Bigg|$$

finding

$$f_Y(y)=\Bigg[\frac{1}{\sqrt{y-1}}-1\Bigg]\mathbb{1}_{(1;2]}(y)$$

this avoid you to calculate the CDF first, as not requested

Observe that: the condition $y \in [1;2]$ is not a restriction but only an useless information; it is the support of Y that you can calculate by your own. In fact trasforming $X \rightarrow Y$ with the transformation $Y=X^2+1$ the original support $X \in [0;1]$ becomes $Y \in [1;2]$.

After calculating $f_Y$ you can see that the support of Y must be $(1;2]$. This doesn't change anything because $P(Y=1)=0$

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It is not forbidden to use the method you wanted to use....but it is longer

1. First you have to derive $F_X(x)=2x-x^2$

2. Second use the definition:

$$F_Y(y)=P(X^2+1 \leq y]=P[X\leq \sqrt{y-1}]=F_X(\sqrt{y-1})=1-y-2\sqrt{y-1}$$

3. Derivating $F_Y(y)$ you get the same result as above.