Suppose we have a IID sample $X_1, X_2, \cdots, X_n$ with each $X_i$ distributed as $\mathcal{N}(\mu, \sigma^2)$. Now suppose we construct (a rather peculiar) estimator for the mean $\mu$: we only choose values from the sample that are greater than a pre-decided value, say $1$, and then take the sample average of only those values:

$$\hat{\mu}=\frac{1}{n_1}\sum\limits_{X_i > 1} X_i$$

Here $n_1$ is the number of values that are greater than $1$. Now I was expecting this estimator to be highly biased. However, we have:

$$\mathbb{E}(\hat{\mu}) = \frac{1}{n_1}\sum\limits_{X_i>1}\mathbb{E}(X_i)=\frac{1}{n_1}n_1\mu=\mu$$

This simply would mean that the estimator is unbiased! But obviously if I do this by generating many numbers and choosing only those which are greater than $1$, I will never get the estimator value less than $1$ and so the estimator has to be biased. What am I missing?

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    $\begingroup$ $n_1$ is a random variable. $\endgroup$
    – Michael M
    Commented Apr 3, 2021 at 6:49
  • $\begingroup$ @MichaelM : Completely missed that. Thanks. How can I correctly calculate the expectation then? $\endgroup$
    – Peaceful
    Commented Apr 3, 2021 at 6:56
  • $\begingroup$ @MichaelM : Probably I can sum the expectations of $\frac{X_i}{n_1}$ but it's not clear to me how to take expectation of the ratio of random variables. $\endgroup$
    – Peaceful
    Commented Apr 3, 2021 at 6:59
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    $\begingroup$ $\mathbb E[X_i]$ is actually $\mathbb E[X_i|X_i>1]$ $\endgroup$
    – Xi'an
    Commented Apr 3, 2021 at 9:00
  • 2
    $\begingroup$ I don't understand why you thought that $\mathbb E(x_i)=n_1\mu$. $\endgroup$ Commented Apr 3, 2021 at 15:39

4 Answers 4


Given that $n_1$ is a random variable (as pointed out already in the comments), the expected value can be computed as $E(\hat\mu)=E_{n_1}[E_{\hat \mu}(\hat\mu|n_1)]$. For the inner expectation, note that one can't just write $E_{\hat \mu}(\hat\mu|n_1)=\frac{1}{n_1}\sum_{X_i>1}E(X_i),$ because the expected value cannot depend on specific values of certain $X_i$, as would be required for the sum. So $$E_{\hat \mu}(\hat\mu|n_1)=\frac{1}{n_1}E\left[\sum_{X_i>1} X_i|n_1\right].$$ For given $n_1$, we can write, with appropriate renumbering of indexes, $\sum_{X_i>1} X_i=\sum_{j=1}^{n_1} X_j^*$, where $X_j^*$ are random variables distributed according to a truncated normal distribution between $a=1$ and $b=\infty$. Let $E_{\mu,\sigma^2,a,b}X$ denote the expectation of such a truncated normal. For $a=1, b=\infty,$ $$E_{\mu,\sigma^2,1,\infty}X=\mu+\frac{\varphi\left(\frac{1-\mu}{\sigma}\right)}{1-\Phi\left(\frac{1-\mu}{\sigma}\right)}\sigma=t>\mu,$$ see https://en.wikipedia.org/wiki/Truncated_normal_distribution . Conditioning on $n_1$, we have $$E_{\hat \mu}(\hat\mu|n_1)=\frac{1}{n_1}\sum_{j=1}^{n_1} E(X_j^*)=\frac{1}{n_1}n_1 E_{\mu,\sigma^2,1,\infty}(X)=t>\mu.$$ This does not depend on $n_1$ (unless $n_1=0$, in which case the sum is empty and $E_{\hat \mu}(\hat\mu|n_1=0)=0$), so ultimately $$E(\hat \mu)=P\{n_1>0\}t.$$ This is $>\mu$ (bias!) if $\mu\le 0$, and also if $P\{n_1=0\}$ is small enough that $P\{n_1>0\}t>\mu$, which should hold unless $n$ is very small (potentially resulting in a large $P\{n_1=0\}$, the value of which is given in Xi'an's solution).

PS: I corrected this seeing Xi'an's solution, who got a thing right that I had forgotten about. That solution is perfectly right as far as I can see, however my different way of getting there may also help.

PPS: I take $\hat \mu=0$ in case $n_1=0$, which isn't entirely clear in the question.


As indicated in my comment (and then in later answers), the error in the reasoning leading to the apparent paradox is to treat the selected or surviving $X_i$'s that we should denote differently, e.g., as $X_j^+$'s, as distributed from the $\mathcal N(\mu,\sigma^2)$ distribution. This accept-reject selection changes the distribution of these $X^+_j$'s to a truncated $\mathcal N^+(\mu,\sigma^2,1)$ distribution, making their expectation equal to $\mathbb E[X_i|X_i>1]$. The number $n_1$ of such $X_j^+$'s is also a Bernoulli random variable (and brings information about $\mu$) but this does not impact the overall expectation (when $n_1\ne 0$).

The estimator $\hat\mu$ need be defined separately when $N_1=0$, for instance setting $\hat \mu(0)=1$ or $\hat \mu(0)=\bar X_n$. Conditional on $N_1=n_1>0$, the selected $X^+_j$'s are thus truncated Normal variates on $(1,\infty)$ with expectation $$\mu+\dfrac{\phi(\sigma^{-1}(1-\mu))}{\Phi(\sigma^{-1}(\mu-1))}\sigma$$ Since $\mathbb P(N_1=0)=\Phi(\sigma^{-1}(1-\mu))^n$, $$\mathbb E[\hat\mu] = \Phi(\sigma^{-1}(1-\mu))^n\hat\mu(0)+[1-\Phi(\sigma^{-1}(1-\mu))^n]\left\{\mu+\dfrac{\phi(\sigma^{-1}(1-\mu))}{\Phi(\sigma^{-1}(\mu-1))}\sigma\right\}$$ and the expectation of $1/N_1$ is not needed.

Note that, for a specific choice of $\hat\mu(0)$ like $\hat\mu(0)=1$, the expectation of $\hat\mu$ may be equal to $\mu$ for some exceptional values of $\mu$, $n$, and $\sigma$, but is biased for almost every parameter value, and definitely so when $\mu<1$. However, $\hat\mu(0)$ can be chosen for the purpose of making $\hat\mu$ unbiased, for instance by using the $X_i$'s (that are all less than $1$ when $n_1=0$).


You're hiding random variables in a few places, then applying linearity of expectation selectively. To avoid these mistakes, rewrite your estimator as

$$ \hat{\mu}=\frac{1}{n_i}\sum_{X_i>1}X_i=\frac{\sum_{i=1}^n X_i \cdot 1[X_i>1]}{\sum_{i=1}^n 1[X_i>1]}, $$

where the random variable $1[X_i>1]$ is 1 when $X_i>1$ and 0 otherwise.

Both the numerator and the denominator are random variables. Furthermore, in the numerator, both $X_i$ and $1[X_i>1]$ are random variables, so the numerator's expectation isn't $N \cdot \mu$.


An intuitive way to compute this $E(\hat{\mu})$ is to consider the following alternative, but equivalent, sampling process.

Step 1: determine $n_1$ independently based on a binomial distribution (with probability $p$ equal to $pr(X>1)$).

Step 2: draw $n_1$ variables $Y_1,\dots,Y_{n_1}$ from a normal distribution truncated on the left at 1.

The variable $\bar{Y} = \frac{1}{n_1}\sum_{i=1}^{n_1} Y_i$ has the same distribution as your $\hat\mu$.

In this example we see that $E(\bar{Y}\vert n_1) = E(Y_\text{truncated})$ independent* from the sample size $n_1$ and so we also have $E(\bar{Y})= E(Y_\text{truncated})$

This means that the expectation value of your $\hat\mu$ does not equal the expectation of the normal distribution but instead the expectation value of a truncated normal distribution.

Thus the estimator is not unbiased. Your equation

$$\mathbb{E}(\hat{\mu}) = \frac{1}{n_1}\sum\limits_{X_i>1}\mathbb{E}(X_i)=\frac{1}{n_1}n_1\mu=\mu$$

Should have been

$$\mathbb{E}(\hat{\mu}) = \frac{1}{n_1}\sum\limits_{X_i>1}\mathbb{E}(X_i|X_i>1)=\frac{1}{n_1}n_1E(Y_\text{truncated}) = E(Y_\text{truncated})$$

The parameter $n_1$ is a variable, as Michael M, mentions in the comments, so it is problematic to eliminate it. But because you get the term $n_1/n_1$ you do not have this problem. The expectation of the average of $n_1$ i.i.d. variables is independent from the sample size $n_1$.

*A tricky thing is what your expression means when $n_1 = 0$. It is undefined in that case because you have a division $0/0$.

  • $\begingroup$ I disagree: Depending on the way $\hat\mu$ is defined when $n_1=0$, there always exists an unbiased version. $\endgroup$
    – Xi'an
    Commented Apr 3, 2021 at 12:01
  • 1
    $\begingroup$ @Xi'an I regard the estimator $\hat\mu$ as undefined for the case $n_1 = 0$. It makes no sense to me to use $\hat\mu = 0$ in case $n_1 = 0$. That would be an arbitrary choice. So you can regard the answer as computing $E(\hat\mu | n_1>0)$. $\endgroup$ Commented Apr 3, 2021 at 13:36

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