I would like to know if it is possible to convert a normal distribution into a triangular distribution. If it is, how it can be done?
I know the mean and the coefficient of variation of the normal distribution.
Solved – Normal distribution to triangular distribution
data transformationdistributionsnormal distributiontriangular distribution
Related Solutions
1) What's depicted appears to be (grouped) continuous data drawn as a bar chart.
You can quite safely conclude that it is not a Poisson distribution.
A Poisson random variable takes values 0, 1, 2, ... and has highest peak at 0 only when the mean is less than 1. It's used for count data; if you drew similar chart of of Poisson data, it could look like the plots below:
$\hspace{1.5cm}$
The first is a Poisson that shows similar skewness to yours. You can see its mean is quite small (around 0.6).
The second is a Poisson that has mean similar (at a very rough guess) to yours. As you see, it looks pretty symmetric.
You can have the skewness or the large mean, but not both at the same time.
2) (i) You cannot make discrete data normal --
With the grouped data, using any monotonic-increasing transformation, you'll move all values in a group to the same place, so the lowest group will still have the highest peak - see the plot below. In the first plot, we move the positions of the x-values to closely match a normal cdf:
In the second plot, we see the probability function after the transform. We can't really achieve anything like normality because it's both discrete and skew; the big jump of the first group will remain a big jump, no matter whether you push it left or right.
(ii) Continuous skewed data might be transformed to look reasonably normal. If you have raw (ungrouped) values and they're not heavily discrete, you can possibly do something, but even then often when people seek to transform their data it's either unnecessary or their underlying problem can be solved a different (generally better) way. Sometimes transformation is a good choice, but it's usually done for not-very-good reasons.
So ... why do you want to transform it?
Using the extreme-order statistics as estimators for the boundaries $a,b$ and then using
$$E(X) = \frac {a+b+c}{3}$$
to estimate $c$ by method of moments is so ...maddeningly easy,
$$\hat a = X_{(1)},\;\; \hat b = X_{(n)},\;\;\hat c = 3\bar X - \hat a - \hat b$$ it made me think how I could start by estimating $c$ first, just for the twist of it. Here it is but not yet with any properties of the estimator. I will make this community wiki in case any one is interested in working it further.
1) Obtain the empirical quartiles $\hat q_1, \hat q_3$ and form the Interquartile Range $\text{IQR} = \hat q_3 - \hat q_1$
2) Use the Friedman-Diaconis rule to bin the data.
$$\text{Bin size}=2\, { \text{IQR} \over{ {n^{1/3}} }}$$
3) Form the empirical histogram and estimate $\hat c$ as the mid-point of the bin with the highest empirical frequency.
4) Then solve for $a,b$ the system of equations
$$q_1 = a + \frac {\sqrt{(c-a)(b-a)}}{2}$$ $$q_3 = b + \frac {\sqrt{(b-c)(b-a)}}{2}$$
using the estimated $\hat q_1, \hat q_3, \hat c$ (the inverse CDF expressions I took from the book chapter the OP links to, page 8).
Best Answer
Yes, it is possible. Basically, what you need is a function $T:{\mathbb R}\rightarrow[a,b]$ such that $F_{a,b}(x)=\Phi_{\mu,\sigma}\left[T\left(x\right)\right]$, where $F_{a,b}$ is the triangular distribution on $[a,b]$, $\Phi_{\mu,\sigma}$ is the normal distribution with mean $\mu$ and variance $\sigma^2$, and $x\in[a,b]$. Then,
$$T\left(x\right)=\Phi^{-1}_{\mu,\sigma}[F_{a,b}(x)].$$
For $\mu=0,\sigma=1,a=-1,b=-1$, $T$ looks as follows
Also, note that the argument used by @COOLSerdash is valid, but the uniform is on the closed interval $[0,1]$, otherwise you cannot transform it into a closed interval $[a,b]$.