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If I have the 10th percentile, mode and 90% percentile for a triangular distribution, how can I find the minimum (i.e. 0th percentile) and maximum (i.e. 100th percentile) of the distribution?

qwertytam
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2 Answers2

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Using wikipedia's parameterization of the Triangular distribution, it should be apparent that $c$ is the Mode of the distribution. The CDF of this distribution is given by:

$$F(x) = \begin{cases} 0 &, x \leq a \\ \frac{(x-a)^2}{(b-a)(c-a)} &, a < x \leq c \\ 1 - \frac{(b-x)^2}{(b-a)(b-c)} &, c < x < b \\ 1 & ,x > b \end{cases}$$

Let's denote the $10^{th}$ and $90^{th}$ percentile of the distribution by $q_{.10}$ and $q_{.90}$ respectively. Then using the CDF we obtain the following system of equations which allows us to solve for $a$ and $b$.

\begin{align*} 0.10 = F(q_{0.10}) \\ 0.90 = F(q_{0.90}) \end{align*}

Example

Suppose $Mode = 1$, $q_{0.10} = 0$ and $q_{0.90} = 2.5$. Clearly we have $c = 1$, and to find $a$ and $b$ we write...

\begin{align*} 0.10 &= \frac{(0-a)^2}{(b-a)(1-a)} \\ 0.90 &= 1 - \frac{(b-2.5)^2}{(b-a)(b-1)} \end{align*}

Get out your pen and paper, or just solve for $a$ and $b$ numerically to obtain:

\begin{align*} a &= -0.93 \\ b &= 3.58 \\ c &= 1 \end{align*}

knrumsey
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  • Thank you. I really appreciate the help. I did start going down the same route. Where I got stuck though, was the algebra to find equations to solve for a and b. The environment where I need to implement this does not have access to a numerical method to solve. Using my long degraded algebra skills, I end up with a very ugly equation! Do you have any hints on how to solve this algebraically? – qwertytam Jul 20 '17 at 13:48
  • With some work, you should be able to get to: $\frac{a^2}{1-a}=\frac{(b-2.5)^2}{b-1}$. You can then write: $a^2 + Ba - B = 0$ where $B = \frac{(b-2.5)^2}{b-1}$. Using the quadratic equation, you should be able to get a (possibly messy) answer for $a$ in terms of $b$. Substitute back into one of the original equations and do your worst. – knrumsey Jul 20 '17 at 18:39
  • So, if we go for the general formula than I have: $$\frac{(q_{0.10}-a)^2}{c-a}=\frac{(b-q_{0.90})^2}{b-c}$$ Using the approach above, then I arrive at: $$a^2 + a(B-2q_{0.10}) + p_{0.10}^2 - Bc = 0$$ Then, with the quadratic equation, I get: $$a = \frac{(2q_{0.10} - B) \pm \sqrt{(B-2q_{0.10})^2 + 4(p_{0.10}^2 - Bc)}}{2}$$ – qwertytam Jul 21 '17 at 02:22
  • Sorry, part 2...and $p_{0.10}$ above should be $q_{0.10}$: So, by rearranging this equation: $$0.10 = \frac{(b-q_{0.90})^2}{(b-a)(b-c)}$$ into terms of a: $$ 0.10a = 0.10b - \frac{(b-q_{0.90})^2}{b-c}$$ It seems like I would end up with a very ugly equation at two points. First when trying to solve the quadratic equation with the substitution of $b$ back into $B$, and second when substituting the quadratic solution into the equation above. Have I made this more complex than necessary? – qwertytam Jul 21 '17 at 02:33
  • Personally, I would just use a numerical approach. If you don't have access to a programming language, even Excel has a "solver" feature that should do the trick. – knrumsey Jul 21 '17 at 17:26
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Another way to describe this is to note that the mode occurs where the line changes from a positive to a negative slope. The mode along with the tenth percentile determines the positive slope and the line can be projected down to determine the minimum where the line hits the x-axis. Similarly the mode and 90th percentile determine the negative slope which can be used to project the line to the maximum where the line once again hits the x-axis. This is what Big Agnes does algebraically. If the triangle is a right triangle there is only one line. If the slope of the line is negative the minimum will be the mode and the maximum can be determine through the negative slope determined using the mode and the 10th (or 90th) percentile and then projecting down to the x-axis. Similarly for a positive slope the maximum is the mode and the slope determined by the mode and the 10th (or 90th) percentile with the line projected down to the minimum when it hits the x-axis. These are degenerate cases.

Michael R. Chernick
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    I don't think this works quite the way you think. Let's take a simple case -- the 10th percentile is 1, the 90th percentile is 11 and the mode is at 8. Unless you're proposing a lengthy exercise involving trial and error, or some algebra you skipped over, I don't see how you're drawing the line on just that information -- because you won't know the height of the density at the peak nor at the 10 and 90th percentiles until you've done some algebraic calculations. If you have a way to avoid that a numerical example (with numbers like those) would be helpful. – Glen_b Jul 20 '17 at 06:41
  • I couldn't follow this answer, perhaps because it does not clearly distinguish the PDF (which, evidently, the "line" refers to) and the CDF, which is piecewise quadratic. – whuber Jul 20 '17 at 22:29
  • I was talking about the triangular shape of the density. As Glen_b remarks you need to determine the height of the mode as well as the 10th and 90th percentile. But one knows that the area between the 10th and 90th percentile is 0.80. It may take trial and error to determine the heights that meet the constraints. – Michael R. Chernick Jul 20 '17 at 22:54