# Find the density function and the distribution function of X [duplicate]

Possible Duplicate:
Functions of Random Variables
Let X be a continuous random variable

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

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## marked as duplicate by Dilip Sarwate, Henry, t.b., Henning Makholm, Asaf KaragilaApr 26 '12 at 7:51

Is this homework? (if so, please add the "homework" tag) In any case, what have you tried and where did you encounter a problem? –  tibL Apr 20 '12 at 14:01
Where does $Y$ come into play? –  David Mitra Apr 20 '12 at 14:03

I will assume that the $X$ at the end is a typo for $Y$. Let us find the cumulative distribution function $F_Y(y)$ of $y$. Then we can differentiate (not done) to find the density function $f_Y(y)$ of $Y$. We have $$F_Y(y)=P(Y\le y)=P((X-1)^2\le y).$$ If $y<0$, this probability is $0$. So $F_Y(y)=0$ if $y<0$. From here on, assume that $y\ge 0$. For $y\ge 0$, we have $$P((X-1)^2\le y)=P(-\sqrt{y}\le X-1\le \sqrt{y})=P(1-\sqrt{y}\le X\le 1+\sqrt{y}). \tag{\ast}$$ Now the analysis breaks down into three cases.
If $y> 9$, the interval in $(\ast)$ fully covers the interval $[-2,2]$, so $F_Y(y)=1$.
Now let $y$ range from $1$ to $9$. For $y$ in this range, $1+\sqrt{y} \ge 2$. So in this range, $P(1-\sqrt{y}\le X\le 1+\sqrt{y})$ is the probability that $X$ ranges from $1-\sqrt{y}$ to $2$. (Remember, the density of $X$ is $0$ beyond $2$.) Thus for $y$ in this range, $F_Y(y)=(2-(1-\sqrt{y}))/4$.
Finally, look at $0 \le y <1$. I leave to you the computation of $F_Y(y)$ for $y$ in this interval.