Change of variables with a continuous but not differentiable mapping

Suppose that Y is a continuous random variable with a density function $f_{Y}(y)$. We transform $Y$ by the following mapping

$$Y^{*} = \left \{ \begin{array}{ll} \alpha Y + \beta & \text{ if } Y < y^{0} \\ Y & \text{ if } Y > y^{0} \end{array} \right .$$ where $\alpha$ and $\beta$ are known constants such that the mapping is continuous, but not differentiable at $Y = y^{0}$. How do I find the density function of $Y^{*}$?

My concern is how the non-differentiable point $Y=y^0$ affects the resulting density of $Y^∗$. My solution for the density of $Y^∗$ is the following:

$f_{Y^{*}}(y^{*}) = f_{Y}(y^{*}) I\{ Y^{*} > y^{0} \} + \frac{1}{\alpha}f_{Y}\left ( \frac{y^{*} - \beta}{\alpha} \right) I\{ Y^{*} < \alpha y^{0} + \beta \}.$

• Is this a problem for a course? If so, please add the [self-study] tag & read its wiki. – gung Oct 6 '14 at 1:15
• It's no a problem for a course. – smileCatcher Oct 6 '14 at 23:03
• My concern is how the non-differentiable point $Y = y^{0}$ affects the resulted density of $Y^{*}$. My solution for the density of $Y^{*}$ is 1. for $Y > y^{0}$ (equivalently, $Y^{*} > y^{0}$), $f_{Y^{*}}(y^{*}) = f_{Y}(y)$; 2. for $Y < y^{0}$, $f_{Y^{*}}(y^{*}) = f_{Y}(y)/\alpha$. – smileCatcher Oct 6 '14 at 23:17
• I tried to add this information into your question. Please ensure it says what you want it to. – gung Oct 6 '14 at 23:23
• A complete and thorough answer to the general question posed here might be difficult to formulate, but in this particular case ask yourself what the probability assigned to the nondifferentiable points is. If that probability is zero, does the lack of differentiability make any difference? – whuber Oct 6 '14 at 23:34