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I have a random variable $K$, given by p. density function $f(x)$: $$ f(x)=\begin{cases} \frac{3}{4}(1-x^2), &|x|<1,\\ \\ 0, &\text{otherwise}\\ \end{cases} $$

I'm trying to calculate the p.density function of a variable $Y=4-K^2$, and at some point I'm stuck: (we know that $-1\leq K\leq 1$): (for $x\in [3,4]$ ) $$ P(Y\leq x)=P(4-K^2\leq x)=P(\sqrt{4-x}\leq K \leq \infty)=\int_{\sqrt{4-x}}^{\infty} f(t)d t=\int_{\sqrt{4-x}}^{1} f(t)d t $$

Now I need to get integral $\int_{0}^{x}f(s)ds$, but I have no clue how to change these variables.

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You are on the right track, with one caveat. You have to remember the negative side of the square root: $$ P(4-K^2 \leq x) = P(\sqrt{4-x} \leq K \leq \infty) + P(-\infty \leq K \leq -\sqrt{4-x}) $$ Conveniently, the probability density function of $K$ is symmetric around zero, so we can say: $$ P(\sqrt{4-x} \leq K \leq \infty) + P(-\infty \leq K \leq -\sqrt{4-x}) = 2P(\sqrt{4-x} \leq K \leq \infty) $$ From there, you are correct, with the additional factor of two, you just have to carry it through: $$ 2\int^{1}_{\sqrt{4-x}}f(t) dt \\ =\frac{3}{2}\int^{1}_{\sqrt{4-x}}1-t^2 dt \\ =\frac{3}{2}\left.\left(t-\frac{t^3}{3}\right)\right|_{\sqrt{4-x}}^1 \\ =\frac{3}{2}\left(\left(1-\frac{1}{3}\right)-\left(\sqrt{4-x}-\frac{(4-x)^{3/2}}{3}\right)\right) \\ =1-\frac{1}{2}\left(x-1\right)\sqrt{4-x} \\ $$ Note that as a sanity check here, you can plug in $x=3$ and $x=4$ and see that the respective probabilities are $0$ and $1$, as we expect.

Differentiate with respect to $x$ to get the pdf.

$$ \frac{d}{dx}1-\frac{1}{2}\left(x-1\right)\sqrt{4-x} \\ =-\frac{1}{2}\left(\sqrt{4-x}+\frac{1-x}{2\sqrt{4-x}}\right) $$

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  • $\begingroup$ .... I am a bit speechless, since I concentrated so hard on trying to manually convert these two integrals I forgot I could just integrate it. Thank you, very much. $\endgroup$ – user19502 Jun 5 '13 at 12:18

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