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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a random variable uniformly distributed on the interval $[−2, 2]$, and $Y = (X − 1)^2$.

$(a)$ Find the density function and the distribution function of $X$.

$(b)$ Find the distribution function and the density function of $Y$.

share|cite|improve this question
Sow us what you have tried: (a) is easy while (b) may require some thought. – Henry Apr 18 '12 at 1:47
Please read: attentively. – t.b. Apr 18 '12 at 1:55
Related meta thread:… – cardinal Apr 19 '12 at 23:57
up vote 1 down vote accepted

I'll leave part a) for you, but I will point out that the density function, $f_X$, for $X$ is zero outside the interval $[-2,2]$; after all, $X$ takes all its values in this interval.

Before tackling part b), it would be beneficial to first determine the values that $Y$ takes: Since $X$ takes values in $[-2,2]$ and since $Y=(X-1)^2$, it follows that $Y$ takes values in the interval $[1,9]$.

From this it follows that $f_Y(x)=0$ for $x\notin[1,9]$. From this it follows that $F_Y(a)=0$ for $a\le1$ and that $F_Y(a)=1$ for $a\ge 9$.

That was the easy part. Let's now find the value of $F_Y(a)$ for $a\in[1,9]$. The idea is write the distribution function of $Y$ in terms of the distribution function of $X$:

We have, for $1\le a\le9$ $$\eqalign{ F_Y(a) &= P[\,Y\le a\,]\cr &=P[\,(X-1)^2\le a\,] \cr &= P [\,1-\sqrt a \le X\le 1+\sqrt a \, ]\cr &= P[\,X\le 1+\sqrt a\,]-P[\,X\le 1-\sqrt a\,]\cr &=F_X(1+\sqrt a) -F_X(1-\sqrt a). } $$

Now you can write the distribution function of $Y$ explicitly: $$ F_Y(a)=\cases{0,\vphantom{1\over2} &$a\le1$\cr F_X(1+\sqrt a) -F_X(1-\sqrt a),\vphantom{1\over2} &$1\le a\le 9$,\cr 1,\vphantom{1\over2} &$a\ge9$ } $$

Of course, in the above, you'd replace $F_X(1-\sqrt a) -F_X(1+\sqrt a)$ with what you obtain from the rule for $F_X$ (note here that $1-\sqrt a$ and $1+\sqrt a$ are in the interval $[-2,2]$).

So, that's $F_Y$. How do you find $f_Y$? As it turns out, this is easy to do now (and is why you were asked to find the distribution function of $Y$ first) by the following result: in general, if $F$ is a distribution function of a continuous random variable, then its derivative gives the corresponding density function of that variable. In our case we have $$ f_Y(x)={d\over dx}F_Y(x). $$ But before we apply this, let's recall we already determined that $f_Y(x)$ is zero for $x$ outside the interval $[1,9]$. For $x\in [1,9]$, we have, using the chain rule and the fact that ${d\over dx} F_X(x)=f_X(x)$, $$\eqalign{ f_Y(x)={d\over dx}F_Y(x) &= {d\over dx}\bigl(F_X(1+\sqrt x) -F_X(1-\sqrt x)\bigr)\cr &={1\over2\sqrt x} f_X(1+\sqrt x) +{1\over2\sqrt x} f_X(1-\sqrt x).\cr } $$ So $$ f_Y(x)=\cases{ {1\over2\sqrt x} f_X(1+\sqrt x) +{1\over2\sqrt x} f_X(1-\sqrt x), &$1\le x\le9$,\cr 0,\vphantom{1\over2}&otherwise . } $$ (And of course you'll want to simplify this using the density of $X$.)

share|cite|improve this answer

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.