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

\begin{equation} Y^{*} = \left \{ \begin{array}{ll} \alpha Y + \beta & \text{ if } Y < y^{0} \\ Y & \text{ if } Y > y^{0} \end{array} \right . \end{equation} 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 \}. $

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    $\begingroup$ Is this a problem for a course? If so, please add the [self-study] tag & read its wiki. $\endgroup$ Oct 6, 2014 at 1:15
  • $\begingroup$ It's no a problem for a course. $\endgroup$ Oct 6, 2014 at 23:03
  • $\begingroup$ 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 $Y < y^{0}$, $f_{Y^{*}}(y^{*}) = f_{Y}(y)/\alpha$. $\endgroup$ Oct 6, 2014 at 23:17
  • $\begingroup$ I tried to add this information into your question. Please ensure it says what you want it to. $\endgroup$ Oct 6, 2014 at 23:23
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    $\begingroup$ 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? $\endgroup$
    – whuber
    Oct 6, 2014 at 23:34

1 Answer 1


The transform leads to a cumulative distribution function that is non-differentiable in the point where you split the transformation.

The density function will have a jump discontinuity in that point. Possibly there are ways to still have the density defined in a way (e.g. use the directional continuity of the cumulative distribution function or an average as you did) but the discontinuity remains.

Below you see an example when we perform such transformation on the normal distribution

example with normal distribution


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