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I am not an expert in statistics but I am facing the following problem: I have a random variable that basically arrises as difference of count data. I have 67 points and they don't seem to be normally distributed. Nevertheless I would like to estimate the probability that the underlying random variable takes values larger than, say, 400.

Would it be statistically sound to estimate a PDF using a kernel density estimation and then compute the probability $P(X>400)$ through that estimation? Or could it be a way to go to fit a normal distribution? However, I have to say that testing on normality fails. Please also see the attached figure.

Thanks. :)

Visualization of the data

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  • $\begingroup$ The difference of two independent Poisson random variables is a Skellam distribution. But if the underlying variables are not Poisson or are not independent, this may not be the appropriate model. en.wikipedia.org/wiki/Skellam_distribution $\endgroup$
    – Sycorax ♦
    Commented Jun 22, 2017 at 14:27
  • $\begingroup$ It is impossible for a "difference of count data" to be normally distributed. You do not need to test for this. $\endgroup$
    – Chris Haug
    Commented Jun 22, 2017 at 22:53
  • $\begingroup$ The two variables I am looking at are highly correlated. In fact the background of my statistical analysis is to show that they have a high correlation, ideally i would have the relation $X_1>X_2$, therefore i won't be able to use the distribution you suggested. $\endgroup$
    – Chris
    Commented Jun 28, 2017 at 10:48

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The probability $P\left(X>t\right)$ you are estimating is the complement of the cumulative distribution function $F(x)=P\left(X\leq t\right)$. which could be nonparametrically estimated by the empirical cumulative distribution function $F_n\left(t\right) = \frac1{n}\sum\limits_{i=1}^n I_{[X_i\leq t]}$. It means that the point estimate of $P\left(X>400\right)$ would just be the number of observations above $400$, which seems to be zero here.

This could be easily supplemented by the confidence interval from the Dvoretzky–Kiefer–Wolfowitz inequality: $$P\left(\sup\limits_t\left|F_n(t)-F(t)\right|>\varepsilon\right)\leq 2e^{-2n\varepsilon^2}$$ Taking the right side to be equal to the desired confidence level $\alpha$, it's easy to show that for any given $t$ the interval $$\left[ 1 - F_n(t) - \sqrt{-\frac1{2n}\log\frac{\alpha}{2}}; 1 - F_n(t) + \sqrt{-\frac1{2n}\log\frac{\alpha}{2}}\right]$$ would cover $P\left(X>t\right)$ with confidence $1-\alpha$ (and it should be trimmed if its boundaries are below $0$ or above $1$).

In your case, with $n=67$ and $F_n(400)=1$ with $95$% confidence we could estimate that the probability of $P(X>400)$ is between $0$ and $\approx0.166$.

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  • $\begingroup$ That's a pretty bad confidence limit. After all, if the probability were well inside it--say, $\Pr(X \gt 400) = 1/10$, then the chance that all $67$ data values are less than or equal to $400$ would be $1 - (1-1/10)^{67}=0.99914\ldots$. A better confidence limit procedure would give a smaller limit, one for which this probability were equal to $1-\alpha$. A much better, and far simpler, limit is thereby obtained as $1-\alpha^{1/n}= 0.0437\ldots$. $\endgroup$
    – whuber ♦
    Commented Jun 22, 2017 at 22:46
  • $\begingroup$ Could you elaborate on how did you get this limit and why does it not depend on the value of the threshold? $\endgroup$
    – Evgeniy Riabenko
    Commented Jun 22, 2017 at 22:56
  • $\begingroup$ It's a Binomial confidence limit. It depends only on the fact that the threshold exceeds the maximum observed value in a simple random sample. See stats.stackexchange.com/search?q=rule+of+three. $\endgroup$
    – whuber ♦
    Commented Jun 23, 2017 at 13:36
  • $\begingroup$ Thanks also for the reference to the rule of three. And I must say the Dvoetzky-Kiefer-Wolfowitz inequality is a really cool result... impressed me from a mathematical standpoint. $\endgroup$
    – Chris
    Commented Jun 30, 2017 at 7:55

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