I understand that any C.D.F may be represented in the form

$$F(x) = p_1F^d(x) + p_2F^c(x)\,,$$

where $F^d(x)$ represents discrete c.d.f , $F^c(x)$ represents continuous c.d.f and $p_1+ p_2=1$.

What is the procedure to decompose the c.d.f in such a form, assuming that it is already a mixed c.d.f?

Consider this cdf :

$$ F(x)= \begin{cases} 0 &,\text{ if }x<0 \\ x^2+0.2 &,\text{ if }0\le x<0.5 \\ x &,\text{ if }0.5\le x<1 \\ 1 &,\text{ if }x\ge 1 \end{cases} $$

How can we express this cdf in the form mentioned above?

  • 2
    $\begingroup$ It depends on how the CDF is given to you. For instance, it could be an algebraic combination of transcendental functions (which is common) or it could be represented indirectly in terms of its PDF, cf, or cgf; or it would be given as a series expansion; or it could even be described in terms of operations on random variables with given distributions. Maybe all you have is a graph! What situation are you in? $\endgroup$
    – whuber
    May 27, 2020 at 16:03
  • $\begingroup$ I have added an example for the c.d.f. I am interested in decomposing only such type of c.d.f's for the time being. Thank you! Also pardon my editing skills, I am not well versed with Latex! $\endgroup$
    – napoleon
    May 28, 2020 at 4:09

1 Answer 1


Note your CDF has two jumps: One of size $\frac{1}{5}$ at $x = 0$ and another of size $\frac{1}{20}$ at $x = \frac{1}{2}$. If we subtract off the jumps at these points we are left with a continuous function,

$$F_c(x) = \begin{cases}0 & x < 0 \\ x^2 & 0 \leq x < \frac{1}{2} \\ x - \frac{1}{4} & \frac{1}{2} \leq x < 1 \\ \frac{3}{4} & x \geq 1 \end{cases} $$

You can think of this is as scaled down continuous CDF (but not a valid one since it does not approach $1$ for $x\rightarrow\infty$). What is the scaling factor, $p_2$ here? In particular, $F_c(x) = p_2*F^c(x)$, where $F^c(x)$ is a valid continuous CDF.

Now consider the jumps we subtracted off earlier. We can create a right-continuous step function using these jumps,

$$F_d(x) = \begin{cases}0 & x < 0 \\ \frac{1}{5} & 0 \leq x < \frac{1}{2} \\ \frac{1}{4} & x \geq \frac{1}{2}\end{cases} $$

You can think of this is as scaled down discrete CDF (but not a valid one since it does not approach $1$ for $x\rightarrow\infty$). What is the scaling factor, $p_1$ here? In particular, $F_d(x) = p_1*F^d(x)$, where $F^d(x)$ is a valid discrete CDF.

Putting things together, you obtain your desired decomposition, $F(x) = p_1*F^d(x) + p_2*F^c(x)$.

  • $\begingroup$ I am having trouble understanding the right reason behind subtracting 1/4 at x=0.5 and x=1 and not 1/20. I know that 0.2 + 0.05=0.25 and also that this a cumulative distribution function but somehow i just can't digest this fact. Can you please explain that too? Also, a general approach to such type of problems would be to find jumps and then remove them to get two different parts? Thanks! $\endgroup$
    – napoleon
    May 28, 2020 at 16:02
  • 1
    $\begingroup$ Sorry, what I said was fairly misleading. First you should subtract off the jump of size 1/5. Then subtract off 1/20 from the linear part of the CDF (ignore the parabola). This should result in one continuous function. $\endgroup$
    – Flowsnake
    May 28, 2020 at 21:04

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