# Finding the domain and distribution of $Y = X^3 + 6$ for $X \sim U[-2,2]$

A small probability problem that I am struggling with...

Let $X \sim U[-2 , 2]$. Find the distribution of $Y = X^3 + 6$.

My main problem is the domain of $Y$.

I found that the domain of $Y$ is $-2 \leq Y \leq 14$, but I believe that the correct one is $6 \leq Y \leq 14$. Also I am a bit confused now, is that $6 \leq Y \leq 14$ actually the domain of $y$ and not $Y$?

Should I rather say that: $Y$'s domain is $[-2,14]$

So $F_Y(y) = P(Y \leq y) = P(x^3+6 \leq y) = P(x \leq \sqrt[3]{y-6})$

So actually $y$'s domain is $[6,14]$?

And then, what integrals should I take? Using what domain?

Thanks a lot!

Thanks all, i think it is sufficiently answered the question.

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This looks like homework, and if so, please add the `homework` tag.  Also, it would help if you used LaTex to clean up your equations.  Now to the problem: Why do you think the domain of $Y$ is $[6,14]$? You found it correctly as $[-2,14]$.  For each real number $y \in [-2,14]$, $$P\{Y \leq y\} = P\{X^3+6 \leq y\} = P\{X\leq (y-6)^{1/3}\}$$ and since $X$ is uniformly distributed, you should be able to compute this probability without needing to integrate. – Dilip Sarwate May 22 '12 at 19:15
Thanks for your answer. Its not that i need to find a probability but i need to find the distribution of Y. So i have to integrate , dont i ? – Peter Mk_dir May 22 '12 at 19:26
For any given real number $y$, the value of the distribution function $F_Y(y)$ at $y$ is a probability: $F_Y(3) = P\{Y \leq 3\}$ is a probability. You will find life a lot easier if you forget all about fancy words and remember instead that for continuous random variables, probabilities are areas under the density curve. $P\{X \leq \sqrt[3]{y-6}\}$ is the area under the density $f_X(x)$ to the left of $\sqrt[3]{y-6}$. Draw a sketch of $f_X(x)$ and see if you can figure out the desired area without writing the symbol $\displaystyle \int$. – Dilip Sarwate May 22 '12 at 19:34

## 2 Answers

What you have written is correct, but why do you say that "So actually $y$'s domain is $[6,14]$?". Note that you are dealing with cube-root (and not square-root) of $(y-6)$ which is defined even when $y-6$ is negative. Just to finish it off, from what you have written we get that $$F_Y(y) = \mathbb{P}(X \leq \sqrt[3]{y-6}) = \begin{cases} 0 & \text{if $\sqrt[3]{y-6} \leq -2$}\\ \frac{\sqrt[3]{y-6} + 2}{4} & \text{if $\sqrt[3]{y-6} \in [-2,2]$}\\ 1 & \text{if $\sqrt[3]{y-6} \geq 2$} \end{cases}$$ Getting rid of the cube-roots in the domain of definition of $F_Y(y)$, we get that $$F_Y(y) = \mathbb{P}(X \leq \sqrt[3]{y-6}) = \begin{cases} 0 & \text{if $y \leq -2$}\\ \frac{\sqrt[3]{y-6} + 2}{4} & \text{if $y \in [-2,14]$}\\ 1 & \text{if $y \geq 14$} \end{cases}$$

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and thanks!!! finally i got it ! – Peter Mk_dir May 22 '12 at 20:00
@PeterMk_dir Kindly accept one of the answers if you have understood and obtained what you want. – user17762 May 22 '12 at 20:17

Let's not worry about words, let's solve the problem. We will find the cumulative distribution function $F_Y(y)$ of the random variable $Y$. This function is, as usual, defined for all reals.

It is clear that if $y<-2$, then $P(Y \le y)=0$. It is also clear that if $y>14$, then $P(Y\le y)=1$. Finally, we deal with $y$ in the interval $[-2,14]$.

We have $Y\le y$ iff $X^3+6\le y$ iff $X\le (y-6)^{1/3}$. There is no problem below $y=6$, since $w^{1/3}$ can be thought of as defined for all $w$, even negative $w$. For $y$ between $2$ and $14$,
$$P(X\le (y-6)^{1/3})=\frac{1}{4}\left((y-6)^{1/3} -(-2)\right).$$
This is directly obtainable from the geometry. However, to deal with more general situations, we observe that $X$ has density function $\frac{1}{4}$ on the interval $[-2,2]$, so the probability is $$\int_{-2}^{(y-6)^{1/3}}\frac{dx}{4}.$$

The conclusion is that $F_Y(y)=0$ if $y<2$, $F_Y(y)= \frac{1}{4}\left((y-6)^{1/3} -(-2)\right)$ if $-2\le y\le 14$, and $F_Y(y)=1$ if $y>2$.

For the density function, differentiate. There is a point of non-differentiability at $y=6$, which one should not worry overly about.

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Very nice explanation, thank you :) – Peter Mk_dir May 22 '12 at 21:49
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