Let $Y$ be a random variable with density $$f_Y(y)=\frac32 y^2$$ if $-1< y <1$; zero otherwise. Find the density of $$U=Y+Y^2.$$
I have done up to $$f_U(u)=\frac{d}{du}F_U(u)=\frac{d}{du}P(Y+Y^2< u),$$
but am not sure how to go from there.
Let Y be a random variable...
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probability-distributions
1 Answers
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Solve the following equation:
$$y^2+y-u=0.\tag 1$$
If there is no solution or there is only one solution then
$$F_U(u)=P(U If there are two solutions $y_1(u) $$F_U(u)=P(Y^2+Y Since the support of the density is the interval $[-1,1]$, the limits of the integration above depends on the location of $y_1(u) $$F_U(u)=P(Y^2+Y List all the possibilities. The following diagram depicting $y^2+y-u$ for different $u$'s will help: It is easy to see that $F_U(u)=0$ if $u\leq-\frac14$ and is $1$ if $u\geq 2$.
For example if $u=0$ then $$F_U(0)=\frac32\int_{-1}^0y^2\ dy=\frac12.$$ The solutions of $(1)$ are $$y_1(u)=\frac12(-1-\sqrt{1+4u}), y_2(u)=\frac12(-1+\sqrt{1+4u}).$$ Combine these results... Then take the derivative of $F_U$ with respect to $u$.