Would someone check my work using cdf method to find the density function?

Question

Problem:

Let Y be a random variable with pdf $$ f_{y}(y) = 2(1-y) \quad 0 < y<1$$ Using the cdf method, find the density function of $U_1 = 2Y-1$.

This seems straightforward, but I would be appreciative if someone would check my work.

My work

Steps:

1. Identify the CDF
2. Take the derivative of said CDF

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Step 1: \begin{align} F_{U_1}(u_1) &= P(U_1 \le u_1) \\ &= P\left(Y \le \dfrac{u_1 + 1}{2}\right) \\ &= \int_{0}^{{(u_{1} + 1)}/{2}} 2(1-y) \space dy \\ &= 2\int_{0}^{{(u_{1} + 1)}/{2}} dy - 2\int_{0}^{{(u_{1} + 1)}/{2}} y \space dy \\ &= 2\bigg[y\bigg]_{0}^{{(u_{1} + 1)}/{2}} - 2 \left[\frac{y^2}{2}\right]_{0}^{{(u_{1} + 1)}/{2}} \end{align}

This should work out to be

$$ F_{U_1}(u_1) = u_1 + 1 - \frac{(u_1 + 1)^2}{2} = \frac{1}{2}(-u^2 + 1) $$

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Step 2:

$ \frac{dF(u_{1})}{du} = \frac{1}{2}(-2u_{1}) = -u_{1}$, so \begin{cases} f_{U_1}(u_1) = -u_1 \quad -1 < u_1 < 1, \\ f_{U_1}(u_1) = 0 \quad \textrm{otherwise} \end{cases}

I think I may have made an error when doing the range. If $ u > 0$ then the value will be less than 0 and cannot be a probability, so my range should go from -1 to 0? Thank you.

Answer 1

The procedure and range are correct, you just made a mistake when calculating the integral. $$ F_{U_1}(u_1) = \int_{0}^{\frac{u_1 + 1}{2}} 2(1-y) dy = \frac{1}{4} (3 + 2u_1 - u_1^2) $$ and $$ f_{U_1}(u_1) = \frac{1}{4}( 2 - 2u_1), \quad -1 < u_1 < 1 $$