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Suppose that $X,Y,Z$ are independent random variables that follow Triangular distribution with $a=1,b=s,c=\frac{s+1}{2}$ ($s > 1$ is some constant).

What is the distribution of $$W=1-\frac{a \cdot \ln(1+|Y-X|)+b \cdot \ln(1+|Z-X|)+c \cdot \ln(1+|Z-Y|)}{(a+b+c) \ln\left(\frac{s-1}{a+b+c}\right)}$$ where $a,b,c$ are some constants?

This seems quite a hard problem. Is there any way to approximate this distribution, use empirical distribution?

I don't need distribution itself, I need to find a value $w$, for which $P(W<w)=p$, $p \in [0,1]$.

Any help will be greatly appreciated.

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    $\begingroup$ If you can sample from X, Y and Z, you can create samples from the target distribution. With a fixed sample of them you can answer a question like yours easily. It is an approximation of course, and depending on how weird the distribution turns out to be you might need a lot of samples. Also depending on how precise your answer needs to be. $\endgroup$
    – Gijs
    Commented Jan 4, 2018 at 11:02
  • $\begingroup$ Actually it appears to be bell-shaped, so I thought it could be normal or t-distribution. $\endgroup$
    – Paul R
    Commented Jan 4, 2018 at 11:03
  • $\begingroup$ Yes, that's possible, sampling is just a very practical approach here. $\endgroup$
    – Gijs
    Commented Jan 4, 2018 at 11:30
  • $\begingroup$ @Gijs It's not possible: this definitely cannot be a Normal or t distribution. ($W$ is an affine transformation of the weighted mean of three bounded nonnegative variables and therefore will be bounded above and below.) Simulation is the fastest and easiest solution for a given $a,s$. That shows these distributions are always positively skewed, but not greatly so. This suggests a general approximation could be developed--perhaps in one form for $s\gg 1$ and another form for $s\approx 1$. A Pearson Type I distribution is a good candidate for approximation. $\endgroup$
    – whuber ♦
    Commented Jan 4, 2018 at 14:19
  • $\begingroup$ @whuber I need to find a critical value, when w is not statistically significant. Something similar to Pearson Correlation coefficient. Is this possible to do something hear. According to empirical probability I've found that P(W<=0.35)=0.95 $\endgroup$
    – Paul R
    Commented Jan 4, 2018 at 18:04

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