Consider a continuous response $Y$ and design matrix vector $\mathbf{X}$. These are related through some function $f(X) = Y$. Suppose that I am interested in estimating the probability that $Y \leq 0.1$ conditional on observing $\mathbf{X}$.

I want to use quantile regression to do this - can I confirm that this is a legitimate methodology?

We have quantiles $\tau \in [0,1]$ and after estimating our quantile regression for each $\tau$ we have our quantile estimates $\mathbf{q} := \{\hat{Q}(\tau) : \tau \in \{0.01,0.02,...,0.99\}\}$. I want to select the $\tau$ such that $\hat{Q}(\tau) \approx 0.1$. When I find such a $\hat{Q}(\tau)$ it seems to then follow naturally that $P(Y \leq 0.1) = \tau$. The reason is that my model has estimated the $\tau$-th quantile to be $0.1$, which is point on the x-axis in $Y$'s pdf that I need to find to be able to determine $P(Y \leq 0.1)$.

In practice this may not work since an estimated quantile can be lower for higher $\tau$ under some $\mathbf{X}$.

Not looking for logistic regression with a discretized response as a solution (since I already know about this).

  • $\begingroup$ A couple of clarifications: 1) Is indexing by $i$ important to the question? 2) You want $P(Y \leq y | X = x)$ but do you know the conditional distribution $P(Y=y|X = x)$? $\endgroup$ Oct 14, 2013 at 16:09
  • $\begingroup$ Why not use logistic regression? You have a binary outcome: greater than 0.1 or not. $\endgroup$
    – zkurtz
    Oct 14, 2013 at 16:35
  • $\begingroup$ @Theja I'm not sure I follow. If I know the latter then I automatically know the former with integration. Also I think the notation is confused -- in the former you use $P$ to mean probability, but in the latter you use it to refer to a density function. $\endgroup$ Oct 14, 2013 at 16:35
  • $\begingroup$ @zkurtz I had this in an old version -- looking for solutions other than logistic regression. $\endgroup$ Oct 14, 2013 at 16:36
  • 1
    $\begingroup$ OK. Maybe it would help clarify your problem if you explain why logistic regression is not ideal here. $\endgroup$
    – zkurtz
    Oct 14, 2013 at 16:38

1 Answer 1


It doesn't appear that $Y$ is binary. Ordinal regression is a good choice here. With any of the ordinal models (proportional odds, proportional hazards, probit, etc.) you can compute the probability that $Y \geq y$ for all $y$. That probability will change at the unique values of $y$. The R rms package orm function implements this efficiently and has a function generator for exceedance probabilities. If you were extremely fortunate and really have Gaussian residuals you can use the maximum likelihood estimator of the exceedance probabilities, which is a simple function of $\hat{\mu}$ and $\hat{\sigma}$.

  • $\begingroup$ Could you please clarify "it doesn't appear that $Y$ is binary". In the question I said it was continuously normally distributed so this comment is a bit confusing to me. $\endgroup$ Oct 14, 2013 at 16:44
  • $\begingroup$ Someone suggested using binary logistic regression but that would not be the best choice if $Y$ is not binary. $\endgroup$ Oct 14, 2013 at 21:39

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