A Question on Random Variable Distribution

Suppose the p.d.f of the random variable X is as follows: $$f(x) = \left\{ \begin{array}{ll} 3x^2 & \mbox{for } 0< x <1 \\ 0 & \mbox{otherwise } \end{array} \right.$$ Also suppose that $$Y = 1 - X^2$$ What is the p.d.f. of Y? I need a detailed answer and if you use the derivative in your solution please tell why you are doing that (The question is from the Probability and Statistics book).

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Our strategy is to first find the cumulative distribution function $F_Y(y)$ of $Y$. This is the time-consuming part.

Then we differentiate $F_Y(y)$ to find the density function $f_Y(y)$ of $Y$. The reason we differentiate is that the density function, for these continuous distributions, is the derivative of the cdf.

Recall that $F_Y(y)=\Pr(Y\le y)$. Since (with probability $1$) $X$ is between $0$ and $1$, $1-X^2$ ranges between $1$ and $0$. We have $$\Pr(Y\le y)=\Pr(1-X^2\le y)=\Pr(X^2\ge 1-y)=\Pr(X\ge \sqrt{1-y}).$$ The last part holds because $X$ only takes on positive values. For random variables $X$ that can take on negative values, things get more complicated.

Now we need to find $\Pr(X \ge \sqrt{1-y})$. Recall that for any $a$ between $0$ and $1$, $$\Pr(X\le a) =\int_0^a 3x^2\,dx=a^3.$$ It follows that $\Pr(X \ge a)=1-a^3$. (We are using here the fact that the probability that $X$ is *exactly $a$ is $0$.)

We conclude that $$F_Y(y)=\Pr(Y\le y)=1-(\sqrt{1-y})^3=1-(1-y)^{3/2}$$ (for $0\lt y\lt 1$). For completeness, we note that $F_Y(y)=0$ if $y\le 0$, and $F_Y(y)=1$ if $y\ge 1$.

Finally, differentiate $F_Y(y)$ to get the density function. We find, using the Chain Rule, that $$f_Y(y)=\frac{3}{2}(1-y)^{1/2}$$ (for $y$ between $0$ and $1$).

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