Let Y be a random variable...

Question

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.

Answer 1

Solve the following equation:

$$y^2+y-u=0.\tag 1$$

If there is no solution or there is only one solution then

$$F_U(u)=P(U<u)=P(Y^2+Y<u)=P(Y^2+Y-u<0)=0.$$

If there are two solutions $y_1(u)<y_2(u)$ (The solutions depend on $u$!) then if $-1\leq y_1(u)<y_2(u)\leq1$ then

$$F_U(u)=P(Y^2+Y<u)=P(Y^2+Y-u<0)=P(y_1<Y<y_2)=\int_{y_1(u)}^{y_2(u)}\frac32y^2\ dy=$$ $$=\frac12\left(y_2(u)^3-y_1(u)^3\right).$$

Since the support of the density is the interval $[-1,1]$, the limits of the integration above depends on the location of $y_1(u)<y_2(u)$. For example if $y_1(u)<-1$ and $y_2(u)>1$ then

$$F_U(u)=P(Y^2+Y<u)=1.$$

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

$$F_U(0)=\frac32\int_{-1}^0y^2\ dy=\frac12.$$

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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 $u$.