Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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 2

up vote 1 down vote accepted

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 $$F_Y(y) = \mathbb{P}(X \leq \sqrt[3]{y-6}) = \begin{cases} 0 & \text{if $y \leq -2$}\\ \frac{\sqrt[3]{y-6} + 2}{4} & \text{if $y \in [-2,14]$}\\ 1 & \text{if $y \geq 14$} \end{cases}$$

share|improve this answer
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.

share|improve this answer
Very nice explanation, thank you :) –  Peter Mk_dir May 22 '12 at 21:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.