Suppose, we have $n$ samples $x_i$ of a random variable: $$x \sim \mathcal N(\mu,\sigma^2) $$

Based on the samples, we want to estimate the probability that $x$ is negative: $$P(x\le0)$$

Intuitively, I would first estimate: $$\hat \mu=\frac 1 n \sum x_i$$ $$\hat \sigma^2={\frac 1 {n-1} \sum (x_i-\hat \mu)^2}$$

and then calculate:

$$P(x\le0)=\frac1{\sqrt{2\pi\hat\sigma^2}} \int_{-\infty}^0 e^\frac{x-\hat\mu}{\hat \sigma} dx$$

However $\hat \mu$ and $\hat \sigma$ have variance! If I use this method, I suspect I am ignoring that variance and making an incorrect estimation.

Is this reasoning right?

If so, how can I estimate $P(x\le0)$ more correctly, taking $\text {VAR}[\hat \mu]$ and $\text {VAR}[\hat \sigma]$ into account?

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    $\begingroup$ Does this help? $\endgroup$
    – David M.
    Commented Feb 25, 2021 at 22:57
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    $\begingroup$ Maximum likelihood estimation is one way (out of many) to do this: see stats.stackexchange.com/questions/492260 (which is exactly the same question, focusing on the MLE solution). $\endgroup$
    – whuber
    Commented Feb 25, 2021 at 23:04
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    $\begingroup$ It all depends on your sample size, if N>100 you may forget about the variance of $\mu$ and $\sigma$, else for small sample sizes, you may use t-test. The link I provided is my question from today and there are more links on additional answers. $\endgroup$ Commented Feb 26, 2021 at 1:26
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    $\begingroup$ Your question is basically asking for how to take the variance/error of point estimates into account while making those estimates. That topic is extremely broad and will create several different types of answers for your specific context. $\endgroup$ Commented Feb 26, 2021 at 11:00

4 Answers 4


The method you are using is very close to the MLE, which has reasonable estimation properties when the underlying parametric model is correct. The MLE has a property called functional invariance, which means that the MLE of a function of the parameters is that function of the MLE. Your method uses the sample variance estimator, which is a bias-corrected version of the MLE of the true variance, but your estimator should have reasonable properties if the underlying model is correct. Of course, you are correct that your estimator involves some variance, but that is true of any estimator in this situation.

If you are confident that your data is from an exchangeable sequence (i.e., it is an IID model) then I would recommend you give serious consideration to instead using the empirical estimator, which is:

$$\widehat{\mathbb{P}(X \leqslant 0)} = \frac{1}{n} \sum_{i=1}^n \mathbb{I}(x_i \leqslant 0).$$

This latter estimator also has good properties, but crucially, it does not rely on the assumption that the data are from a normal distribution. The empirical estimator is consistent for any underlying distribution (which your estimator is not) which makes it highly robust to model misspecification.

  • $\begingroup$ thank you! sorry what is that $\mathbb{I}$ operator? $\endgroup$
    – elemolotiv
    Commented Feb 26, 2021 at 6:54
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    $\begingroup$ I use that to denote the indicator function for an event. So $\mathbb{I}(x_i \leqslant 0)$ is equal to one if $x_i \leqslant 0$ and zero otherwise. $\endgroup$
    – Ben
    Commented Feb 26, 2021 at 7:32
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    $\begingroup$ For small $n$ or small $P(X \geq 0)$ this estimator is gonna have a much higher variance such that the robustness may be less of an improvement. $\endgroup$ Commented Feb 26, 2021 at 11:07
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    $\begingroup$ I've computed this variance now for the MLE in this question stats.stackexchange.com/questions/511359 Based on that we can say that the variance of this estimator, relative to the variance of the MLE based estimator is approximately $$\frac{\Phi(-d)}{(1+\frac{1}{2}d^2)\phi(d)^2}$$ where $d = \mu/\sigma$. This factor is about twice higher for $d=1$ and increases for smaller or higher $d$. The variance decreases proportional to $\frac{1}{n}$, for small $n$ the variance might be more important than the bias. $\endgroup$ Commented Feb 26, 2021 at 13:52
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    $\begingroup$ Of course it does; if the underlying distribution is normal then you get an advantage from working that into the estimator. However, try simulating the estimates for, say, a highly skewed or highly kurtotic distribution, and you'll see that the empirical estimator will massively outperform the MLE (the latter won't even be consistent). That is the robustness advantage of the empirical estimator. $\endgroup$
    – Ben
    Commented Feb 26, 2021 at 21:40

You can do better.

In the sense that among all the unbiased estimators, you can find the one with the smallest variance.

Our goal is to estimate $\mathbb{P}(X<0)$. An unbiased estimator is $$\mathbb{1}(X_1<0)$$ because $\mathbb{E}[\mathbb{1}(X_1<0)]=\mathbb{P}(X_1<0)=\mathbb{P}(X<0)$.

Since this is unbiased, from Lehmann–Scheffé theorem, we know that $$\mathbb{P}(X_1<0|\bar{X},S)=\mathbb{P}(X_1<0|\hat{\mu},\hat{\sigma}^2)$$is the uniform-minimum variance unbiased estimator or UMVUE, because $(\bar{X},S)$ is complete and sufficient.

A few more step. The above equals to $$\mathbb{P}(\frac{X_1-\bar{X}}{S}<\frac{-\bar{X}}{S}|\bar{X},S)$$ By Basu's thorem, we know $\frac{X_1-\bar{X}}{S}$ is independent with $(\bar{X}, S)$, because $\frac{X_1-\bar{X}}{S}$ is ancillary and $(\bar{X}, S)$ is a sufficient complete statistic. This means the above equation is nothing but $$\mathbb{P}(\frac{X_1-\bar{X}}{S}<\frac{-\bar{X}}{S})$$

Now, all we are left to do is to find out the distribution of random variable $T=\frac{X_1-\bar{X}}{S}$, which is given precisely here.

For more information, please see "Theory of Point Estimation SECOND EDITION Lehmann & Casella", example 2.2.2.

------------------- a comment on your estimator ------------------

Your estimator is good enough in the sense that it is consistent, namely, when you have a large sample size, your estimator will be very close to the true value. This is because of the consistency of the plug-in method, or it can be shown by the continuous mapping theorem. However, it is a biased estimator.

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    $\begingroup$ @SextusEmpiricus It is the first observation. As you can see it has been conditioned on a complete sufficient statistic, which would then give the UMVUE. For the distribution of $T$, one can refer to stats.stackexchange.com/q/181964/119261. $\endgroup$ Commented Feb 26, 2021 at 13:48
  • $\begingroup$ @StubbornAtom I see I have misread this answer. In "An unbiased estimator is $\mathbb{1}(X_1<0)$" I read that this is the UMVUE. $\endgroup$ Commented Feb 26, 2021 at 14:11

Suppoxe $X \sim \mathsf{Norm}(\mu=1,\sigma=10).$ then $P(X < 0\,|\,\mu=1, \sigma=10) = 0.4602.$

pnorm(0, 1, 10)
[1] 0.4601722

Now suppose you have a random sample x of size $n = 50$ from $\mathsf{Norm}(\mu=1,\sigma=10).$

x = rnorm(50, 1, 10)
mean(x);  sd(x)
[1] 2.747168
[1] 10.84025

Then the usual estimates are $\hat\mu=\bar X =2.7472,$ and $\hat\sigma = S = 10.84025.$ Accordingly, you suggest $\hat P(X < 0\,|\,\hat\mu,\hat\sigma) = 0.400.$

pnorm(0, mean(x), sd(x))
[1] 0.3999707

There are various ways of assessing the variability of this estimate of the probability. One possibility is to give a 95% parametric bootstrap confidence interval of the probability. There are many styles of bootstraps and bootstrapping is not the only possibility.

To get the discussion started, the simple quantile bootstrap method shown below gives the interval $(0.286, 0.511),$ centered near our point estimate $0.40.$ Because this is a simulation procedure (starting with known $\mu$ and $\sigma$ to get the data for estimation), we know that the true probability is $0.46,$ and thus that this CI contains the true probability. However, in an actual application we would not know whether such a bootstrap CI covers (contains) the true probability; we can hope that, for 95% of samples of size $n=50,$ it does.

a = mean(x);  s = sd(x)  # sample 'x' from above
B = 3000;  p.re = numeric(B)
for(i in 1:B) {
  x.re = rnorm(50, a, s)
  p.re[i] = pnorm(0, mean(x.re), sd(x.re))
quantile(p.re, c(.025,.975))
      2.5%     97.5% 
0.2861269 0.5109705 

[1] 3000

Here is a histogram of the 3000 uniquely different re-sampled probability estimates used in making the bootstrap. Sometimes correction for skewness in bootstrap CIs is warranted, but our bootstrap distribution seems roughly symmetrical.

hist(p.re, prob=T, br=20, col="skyblue2")
 abline(v = c(0.286, 0.511), col="red", lwd=2, lty="dotted")

enter image description here

I will be interested to see other ideas how to approach this problem.

  • $\begingroup$ thanks! so $P(x\le0 | x_1 ... x_n)$ has itself a probability distribution, right? $\endgroup$
    – elemolotiv
    Commented Feb 26, 2021 at 7:21
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    $\begingroup$ @elemolotiv No, unless you're a Bayesian. That's a statistic. $\endgroup$
    – AdamO
    Commented Feb 26, 2021 at 15:25

The variance of your estimates doesn't matter. The mean and variance estimates are consistent, so the functional representation is consistent as well. We use the degree of freedom correction for the estimate of the sample variance because it's oh-so slightly biased, but that bias disappears in moderate samples. Note: the biased estimator (without the degrees of freedom correction) is the maximum likelihood estimator.

MLEs and functions of them are consistent estimators of the parameters and functions of the parameters respectively a result of the continuous mapping theorem. When the normal assumption holds, your proposed estimator is better than the non-parametric estimate (the proportion of $X$ that is negative). It likely has lower variance too. To find the variance, you can use the law of total variance, or the delta-method.


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