# [R] Dequantizing

Richard.Cotton at hsl.gov.uk Richard.Cotton at hsl.gov.uk
Thu Nov 20 18:23:45 CET 2008

```> I have some data measured with a coarsely-quantized clock.  Let's say
> the real data are
>
>       q<- sort(rexp(100,.5))
>
> The quantized form is floor(q), so a simple quantile plot of one
> against the other can be calculated using:
>
>       plot(q,type="l"); points(floor(q),col="red")
>
> which of course shows the characteristic stair-step.  I would like to
> smooth the quantized form back into an approximation of the underlying
> data.
>
> The simplest approach I can think of adds a uniform random variable of
> the size of the quantization:
>
>       plot(q,type="l"); points(floor(q),col="red");
> points(floor(q)+runif(100,0,1),col="blue")
>
> This gives pretty good results for uniform distributions, but less
> good for others (like exponential).  Is there a better
> interpolation/smoothing function

I'm not convinced that adding a random amount to the floor values to 'make
up' the underlying data is very meaningful.

If you know what the underlying distribution is, then you are best off
using this distribution to generate plots and extra pretend data.

If you know the distribution is exponential, then you can estimate the
rate by treating the true values as interval censored data, somewhere
between floor and floor+1.

library(survival)
q <- sort(rexp(100,.5))
qq <- floor(q)
qq[qq==0] <- 0.00001     #survreg doesn't like values that are exactly
zero
ss <- Surv(qq, qq+1, type="interval2")
model <- survreg(ss ~ 1, dist="exponential")
summary(model)
rate <- 1/exp(model\$coefficients["(Intercept)"]); rate   #hopefully
something close to 0.5

If you don't know the underlying distribution either, then things get
trickier.  Try a histogram/kernel density plot/boxplot to see what it
looks like.

Regards,
Richie.

Mathematical Sciences Unit
HSL

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