# Deriving a distribution whose pdf has the shape of a square + a triangle (a right trapezoid)

I want to the derive the PDF which looks like the sum of a triangular and uniform distribution which looks like this:

To do this I have simply added the PDFs for the rectangular and triangular parts, over the range $$[n,N].$$

$$f(x) = \frac{2(N-x)}{(N-n)^2}$$

$$g(x) = \frac{1}{N-n}$$

Then (I believe), the compound distribution is simply:

$$h(x) := f(x) + g(x) = \frac{3N -2x -n}{(N-n)^2}$$

However, I do get a bit confused here, since this distribution needs to be normalised, which is simply done as so:

$$h_{\text{norm}}(x) = \frac{1}{\int_x h(x)} h(x)$$

Does this seem reasonable, or am I wildly off-chart here?

This is a related question but it seems very complicated, for what should be quite simple.

• Could you explain what you mean by a "compound pdf"? What is it intended to represent? Your picture is puzzling because the areas of the square and triangle differ, demonstrating their graphs cannot both represent densities.
– whuber
Dec 29, 2018 at 18:17
• The compound part is the combination of the square and the triangle. Imagine that you know the PDF of each individual geometry and you would like to find an expression for the combination of both PDFs, that is what I mean by compound. I am studying a problem which gives rise to a histogram of this shape, this is the reason for me wanting to find an analytical expression for the distribution. Dec 29, 2018 at 21:43
• Could you explain what "the PDF of [an] individual geometry" might possibly mean? And exactly how one "combines" two PDFs (and how that would be interpreted)? I am concerned that your question might ultimately be based on a misunderstanding of what these terms mean in a particular application. If the concern merely is of a histogram of this shape, then that is thoroughly addressed in the related question you identified, suggesting your interest is a little broader (or different) than that.
– whuber
Dec 29, 2018 at 22:56
• It is fully possible that I have misunderstood it. Though @jbowman has provided a result below which is what I was broadly looking for. The derived result below is also less complex than the one I linked. When I say the PDF of an individual geometry it could be e.g. the triangular distribution (PDF of a triangle) or a trapezoidal distribution (PDF of a trapezium). I also fully admit that I may have abused terminology here and would be happy to be corrected. Ultimately I was just interested in finding the PDF of the area under the sloped line in the above figure. Dec 29, 2018 at 23:00
• @Astrid could you clarify whether you mean compound distribution or mixture distribution. Also, is the expression for the triangular distribution (which is more like a trapezium) what you want? Jan 18, 2019 at 9:33

The first step is to find an equation for the unnormalized density function, which in this case is the line at the top of your graph:

$$f(x) \propto 9 - {4(x-n) \over N-n}$$

We then integrate this over the range $$[n,N]$$ to find the constant of integration $$c$$:

$$c = \left(9 + {4n \over N-n}\right)\int_n^Ndx \quad - \quad {4 \over N-n}\int_n^Nxdx$$

Working through the integrals gets us to:

$$c = 9N - 9n + 4n -2(N-n)$$

which simplifies to $$c=7N-3n$$. Combining this with our unnormalized density function and rearranging terms leads to:

$$f(x) = {9N - 5n -4x \over (7N - 3n)(N-n)}$$

• For clarity; the result $f(x)$ is not a proper density since it is not normalised yet? I.e. finding the constant of integration does not automatically yield us a normalised density. Further, I calculate the slope as $m=\frac{5-9}{N-n}$ but you have somehow got an extra $-n$ in there -- where did it come from? Also thanks, this is great. Dec 29, 2018 at 22:05
• 1. The final result is normalized, as we divided by $(7N-3n)$, the constant of integration. 2. The slope isn't all we want; we want the value of $f(x)$ for each $x$, and that requires knowing the intercept too. You'll observe that the slope as I have it is $-4/(N-n)$, the same as what you have found, but the intercept does depend on $n$, and that's where the extra $4n/(N-n)$ comes from. Dec 29, 2018 at 22:35
• jbowman, to get the CDF of this expression I simply integrate $f(x) = {9N - 5n -4x \over (7N - 3n)(N-n)}$ once if I have understood the process correctly? Dec 30, 2018 at 15:16
• That's right, and expected values etc. are done similarly. Dec 30, 2018 at 15:26
• @jbowman It disturbed me how this outcome was different from mine. I see now where the difference is. You evaluated $\int_n^N x dx = \frac{1}{2} (N-n)^2$ missing a part $n(N-n)$. Then this will add an extra term $-4n$ and it should be $c=7N-7n$ instead of $c=7N-3n$ Jan 18, 2019 at 11:04

Your image shows the sum of two functions which relates to a mixture distribution:

$$h(x) = a g(x) + (1-a) f(x)$$

with

• the continuous distribution:

$$g(x) = \begin{cases} \frac{1}{N-n} & \quad \text{ for n \leq x\leq N } \\ 0 & \quad \text{otherwise}\end{cases}$$

• a triangular distribution:

$$f(x) = \begin{cases} 2 \frac{N-x}{(N-n)^2} & \quad \text{ for n \leq x\leq N } \\ 0 & \quad \text{otherwise}\end{cases}$$

You do not need to worry about the constant of integration since:

$$\begin{array}{rcl} \int_n^N h(x)dx &=& \int_n^N \underbrace{( a g(x) + (1-a) f(x))}_{=h(x)} dx \\ & = & \int_n^N a g(x) dx + \int_n^N (1-a) f(x) dx \\ & = & a \underbrace{\int_n^N g(x) dx}_{=1} + (1-a) \underbrace{\int_n^N f(x) dx}_{=1} \\ & = & a + (1-a) = 1 \end{array}$$

To get your figure you need to add 5/7 times the uniform (rectangular) distribution and 2/7 times the triangle distribution.

$$h(x) = \frac{5}{7} g(x) + \frac{2}{7} f(x) = \begin{cases} \frac{\frac{5}{7} + \frac{4}{7} \frac{N-x}{N-n} }{N-n} & \quad \text{ for n \leq x\leq N } \\ 0 & \quad \text{otherwise}\end{cases}$$