# Examples of distributions with easily solvable quantile functions but hard to solve CDFs

I'm interested in examples of probability distributions where the quantile function $$F^{-1}(p)$$ exists in closed form or is easy to calculate but where the cumulative distribution function (CDF) $$F(x)$$ does not exist in closed form. In particular examples where the quantile function is easy to evaluate but the CDF is difficult or impossible to solve interest me, especially when the distribution is used in real world applications. Any example would be much appreciated!

• I dont understand the downvote! I believe there are examples on site, will search when some more time Commented Jul 19 at 23:30
• Possibly this should be community wiki? Any function on the domain [0,1] that is difficult to invert is an answer. Commented Jul 21 at 19:08

For a book-length answer to your question, see Statistical Modelling with Quantile Functions by Warren Gilchrist. In the following we assume random variables are continuous, let $$F$$ the the cdf of $$X$$, then the quantile function is the inverse of $$F$$, $$Q(p) = F^{-1}(p), p \in [0, 1]$$ $$Q$$ is then a monotonically increasing function, and importantly, any monotonically increasing function of argument $$p\in [0,1]$$ is the quantile function of some (real) random variable. That makes it easy to construct new quantile function from existing ones, making for flexibility in modeling, and we can use that to easily make some quantile functions that do not have analytical inverses.

Some examples:
Sums of quantile functions are quantile functions
Sums of monotonically increasing functions are also monotonically increasing

Quantile functions to a positive power
A positive power $$Q^\alpha$$ of an increasing function is increasing

Convex combinations of quantile functions
If $$Q_1, Q_2$$ are increasing, then so is $$\omega Q_1 + (1-\omega) Q_2$$ for $$0\le \omega \le 1$$

Positive linear combinations of quantile functions
If $$Q_1, Q_2$$ are increasing, so is $$a Q_1 + b Q_2$$ for $$a\ge 0, b\ge 0$$

Products of quantile functions corresponding to non-negative random variables
If $$Q_1 \ge 0, Q_2 \ge 0$$ then so is $$Q_1 Q_2 \ge 0$$. This also works if only one of the quantile functions correspond to a non-negative random variable.

(Some of this conditions can be summarized by saying that the set of quantile functions constitute a convex cone)

So for some examples we can construct using this.

The (standard) exponential distribution have quantile function $$Q(p)=-\log(1-p)$$. If we reflect this distribution in the $$y$$-axis, the quantile function becomes $$\log p$$. The sum of this quantile functions is $$\log(p) - \log(1-p) = \log\left( \frac{p}{1-p} \right)$$ which is the quantile function of the logistic distribution. If we take a convex combination, $$\alpha \log p - (1-\alpha) \log(1-p)$$ we get a skew distribution, the skew logistic, which do not have an analytic cdf.

The quantile function of the standard uniform random variable is $$Q(p)=p$$. We can combine this linearly with the logistic to get $$Q(p)= \log\left( \frac{p}{1-p} \right) + Kp, K>0$$ which will have the effect of flattening the maximum. This quantile function can not be inverted analytically, either. Now you have the tools to construct many examples yourself!

(I will add some plots later)

• This is a great answer. Commented Jul 21 at 5:57
• I like your use of monotonously increasing (perhaps suggested by a spellchecker) but I think monotonically increasing may be more common. Commented Jul 22 at 9:11
• Thanks a lot for this answer! Commented Jul 22 at 13:35
• "Monotonic" means always the same (unchanging; in the same direction); "monotonous" means boring ;-).
– whuber
Commented Jul 24 at 20:47
• @whuber: Thanks, will edit ... not a native english speaker (or writer) Commented Jul 24 at 23:11

Assuming you mean evaluate rather than solve*, the Tukey lambda distributions have easily evaluated quantile functions but the cdf doesn't have closed form

https://en.wikipedia.org/wiki/Tukey_lambda_distribution

Among other uses, particular members of the family were somewhat commonly used in the past to produce approximate normal scores.

* Let me explain why I am making a deal about whether you mean evaluate rather than solve. The distinction between evaluate and solve is central to the question being answered. It's not just me pedantically insisting on conventionally-correct terminology in this instance; changing from evaluate to solve quite literally flips us from the easy case to the hard case and vice versa:

While $$F^{-1}(p)$$ is readily calculated by direct substitution, ('easy' per your conditions), if I set $$F^{-1}(p) = x_0$$ and try to solve that equation to obtain $$p$$, it is hard. Indeed, it's equivalent to the task of evaluating $$F$$ itself. If we could solve that equation easily, we could evaluate $$F$$ easily. Typically you'd need to use some iterative root finding method or some series approximation etc, which is not nearly so straightforward.

• I see what you mean. I did mean evaluate. I'll update the question! Commented Jul 22 at 13:28

This situation can happen whenever there is matching. For instance:

• Suppose unmarried men’s wealth and unmarried women’s wealth both have Pareto distributions, with quantile functions $$m (1-p)^{-1/\alpha}$$ and $$w(1-p)^{-1/\beta}$$. If men and women pair off according to wealth, then the quantile function for couple’s wealth is $$m (1-p)^{-1/\alpha}+w(1-p)^{-1/\beta}$$ which almost always has no closed-form cdf.

• Suppose a process requires one step with a uniform distribution of times and one step with an exponential distribution of times, and the same processes get quicker times on both. Then the individual quantile functions might be $$ap$$ and $$b\log(1/(1-p))$$, while the combined quantile is $$a p+b \log(1/(1-p))$$ which has no closed-form cdf.