Question 1170932: Let Y = X^2 where X is U(a,b), 0<=a
Show that the pdf of Y is f(y) - 1/[2sqrt(y)(b-a)] for a^2<=y <= b^2.
I know to start I need x = sqrt(y) and to plug into the formula f(g(y))du/dy to find the pdf but I am struggling from there.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's derive the probability density function (PDF) of $Y = X^2$, where $X$ is uniformly distributed on the interval $[a, b]$, with $0 \le a < b$.
**1. PDF of X**
Since $X$ is uniformly distributed on $[a, b]$, its PDF is:
$$f_X(x) = \begin{cases} \frac{1}{b-a} & \text{for } a \le x \le b \\ 0 & \text{otherwise} \end{cases}$$
**2. Cumulative Distribution Function (CDF) of Y**
We want to find $F_Y(y) = P(Y \le y)$. Since $Y = X^2$, we have:
$$F_Y(y) = P(X^2 \le y) = P(-\sqrt{y} \le X \le \sqrt{y})$$
Since $a \ge 0$, $X$ is non-negative, so we only need to consider the positive root:
$$F_Y(y) = P(a \le X \le \sqrt{y})$$
Now, we can express this in terms of the CDF of $X$:
$$F_Y(y) = F_X(\sqrt{y}) - F_X(a)$$
The CDF of $X$ is:
$$F_X(x) = \begin{cases} 0 & \text{for } x < a \\ \frac{x-a}{b-a} & \text{for } a \le x \le b \\ 1 & \text{for } x > b \end{cases}$$
So,
$$F_Y(y) = \frac{\sqrt{y} - a}{b-a} - \frac{a-a}{b-a} = \frac{\sqrt{y} - a}{b-a}$$
This is valid for $a^2 \le y \le b^2$.
**3. PDF of Y**
To find the PDF of $Y$, we differentiate the CDF of $Y$ with respect to $y$:
$$f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{d}{dy} \left( \frac{\sqrt{y} - a}{b-a} \right)$$
$$f_Y(y) = \frac{1}{b-a} \frac{d}{dy} (\sqrt{y} - a) = \frac{1}{b-a} \frac{d}{dy} (y^{1/2} - a)$$
$$f_Y(y) = \frac{1}{b-a} \left( \frac{1}{2} y^{-1/2} - 0 \right) = \frac{1}{2(b-a)\sqrt{y}}$$
Therefore, the PDF of $Y$ is:
$$f_Y(y) = \frac{1}{2\sqrt{y}(b-a)} \quad \text{for } a^2 \le y \le b^2$$
And $f_Y(y) = 0$ otherwise.
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