# Is the sample quantile unbiased for the true quantile?

I would like to find a way to show whether the sample quantile is an unbiased estimator of the true quantiles. Let $$F$$ be strictly increasing with density function $$f$$. I will define the $$p$$-th quantile for $$0 as $$Q(p)=F^{-1}(p)$$ and the sample quantile as $$\hat{F}_n^{-1}(p)=\inf\{x:\hat{F}_n(x)\geq p\},$$ where $$\hat{F}_n(x)$$ is the empirical distribution function, given by $$\hat{F}_n(x)=\frac{1}{n}\sum_{i=1}^n I(X_i \leq x).$$ Based on literature I have read, I expect the sample quantile to be biased, but I am having trouble figuring out how to take the expected value of $$\hat{F}_n^{-1}(p)$$, particularly since it is defined as the infimum of a set. I do know that the expected value of the empirical distribution function is $$F(x)$$. Any help or references that could guide me would be greatly appreciated!

• Cross-posted at math.stackexchange.com/q/3378799/321264. – StubbornAtom Oct 3 '19 at 4:39
• Asymptotically unbiased for $p$th sample quantile (other than max and min), of a continuous distributions with positive density at $p$th population quantile. Asymptotic dist'n of the sample quantile is normally distributed with pop quantile as mean. (Sort of a CLT for 'central' quantiles.) – BruceET Oct 5 '19 at 3:54

Assuming that $$X_1, X_2, X_3 \sim \text{IID } F$$ the empirical distribution function has a scaled binomial distribution:

$$\hat{F}_n(x) \sim \frac{1}{n} \cdot \text{Bin}(n, F(x)).$$

For a given probability value $$0 < p < 1$$ we will denote the sample quantile as:

$$\hat{Q} \equiv \hat{Q}_n(p) \equiv \inf \{ x \in \mathbb{R} | \hat{F}_n(x) \geqslant p \}.$$

Since the empirical distribution function $$\hat{F}_n$$ is non-decreasing and right-continuous, we have the event equivalence $$\inf \{ x \in \mathbb{R} | \hat{F}_n(x) \geqslant p \} \leqslant q$$ if and only if $$\hat{F}_n(q) \geqslant p$$. Thus, the distribution function for the sample quantile is:

\begin{equation} \begin{aligned} F_{\hat{Q}}(q) = \mathbb{P}(\hat{Q} \leqslant q) = \mathbb{P} \bigg( \inf \{ x \in \mathbb{R} | \hat{F}_n(x) \geqslant p \} \leqslant q \bigg) = \mathbb{P} \big( \hat{F}_n(q) \geqslant p \big). \\[6pt] \end{aligned} \end{equation}

In order to look at the bias of the sample quantile as an estimator of the true quantile, we need to look at the expected value of the former. Using a general expectation rule shown here, the exact expected value of this random variable can be written as the integral:

$$\mathbb{E}(\hat{Q}) = \int \limits_{-\infty}^\infty \Big[ \mathbb{I}(q \geqslant 0) - F_{\hat{Q}}(q) \Big] dq = \int \limits_{-\infty}^\infty \Big[ \mathbb{I}(q \geqslant 0) - \mathbb{P} ( \hat{F}_n(q) \geqslant p ) \Big] dq.$$

This integral is complicated, owing to the scaled binomial distribution for $$\hat{F}_n$$. However, as $$n \rightarrow \infty$$ we have $$\hat{F}_n(q) \rightarrow F(q)$$, and so if $$F$$ is continuous at $$q$$ then we also have $$Q(\hat{F}_n(q)) \rightarrow q$$. This gives the asymptotic convergence:

$$\mathbb{E}(\hat{Q}) \rightarrow \int \limits_{-\infty}^\infty \Big[ \mathbb{I}(q \geqslant 0) - \mathbb{I} ( q \geqslant Q(p) ) \Big] dq = \int \limits_{0}^{Q(p)} dq = Q(p),$$

so long as $$F$$ is continuous at $$p$$. Thus, you should expect the sample quantiles to be asymptotically unbiased, except at quantiles corresponding to points of discontinuity of the underlying distribution function. Obviously we may have non-zero bias for finite samples, with the bias depending on the form of the underlying distribution.

• (+1) Right about bias for small samples. Simulation in R for $10^6$ samples of size $n=5$ from $\mathsf{Exp}(1):$ h = replicate(10^6, median(rexp(5, 1))); mean(h) returns 0.783058, but population median is 0.6931472. – BruceET Oct 5 '19 at 4:03
• @BruceET: an exception being the median of a symmetric distribution: for instance, mean(replicate(1e6,median(rnorm(5)))) returns -0.001093016 when mean(replicate(1e6,mean(rnorm(5)))) returns -0.0001424392. – Xi'an Apr 29 '20 at 14:01

I do not think Ben's derivation is completely correct. The asymptotic unbiasedness feature of sample quantile is not distribution free. There is an important assumption that the r.v. has to satisfy: there is a unique solution $$x$$ to the condition $$F(x-) \leq p \leq F(x)$$. A counter example:

Let $$X_1, \dots, X_n \sim$$ $$X$$ i.i.d. where $$X$$ is Bernoulli taking $$-1$$ with probability $$0.5$$ and $$1$$ with probability $$0.5$$. Now let $$p = 0.5$$, then the theoretical quantile (median) $$Q(p)$$ should be $$-1$$. Now for the sample quantile $$Q_n(p) = \begin{cases} -1, & \sum_{i=1}^n\mathbf{1}_{X_i = -1} \geq n/2 \\ 1, & \sum_{i=1}^n\mathbf{1}_{X_i = -1} < n/2. \end{cases}$$ Therefore, $$E(Q_n(p)) = -P\left(\frac{\sum_{i=1}^n\mathbf{1}_{X_i = -1}}{n} \geq 1/2\right) + P\left(\frac{\sum_{i=1}^n\mathbf{1}_{X_i = -1}}{n} < 1/2\right).$$ Now if we take the limit, because of CLT, $$\lim_{n\to\infty}P\left(\frac{\sum_{i=1}^n\mathbf{1}_{X_i = -1}}{n} \geq 1/2\right) = \lim_{n\to\infty}P\left(\frac{\sum_{i=1}^n\mathbf{1}_{X_i = -1}}{n} < 1/2\right) = 0.5.$$ Therefore, $$\lim_{n\to\infty}E(Q_n(p)) = 0 \neq -1$$.

In fact, we can also check this with mean(2*(replicate(2000, mean(2*rbernoulli(10^6)-1 ==-1)) >= 0.5)-1) which gave an answer of -0.011. The issue is when the solution to $$F(x-) \leq p \leq F(x)$$ is not unique, i.e., there are segments where the cdf are flat. The sample quantile will jump around and does not settle down.

• Well-spotted (+1) --- I have amended my answer to add the required continuity condition. – Ben Oct 5 '20 at 20:21
• Actually I shouldn't say continuous (I corrected it). It should be there is only a unique solution $x$ to the condition $F(x-) \leq p \leq F(x)$. Because you can also have $X\sim unif(0,1)\cup(2,3)$ and pick $p = 0.5$. This is similar to the counterexample I gave. The true median is 1 but the asymptotic expectiation of sample quantile go to $1.5$. But $X$ here still has continuous cdf everywhere. – NamelessGods Oct 5 '20 at 21:57
• @NamelessGods In your comment example, I would say the median is any value from $1$ through to $2$, including $1.5$, and in your main example the median is any value from $-1$ through to $+1$, including $0$ – Henry Oct 6 '20 at 8:13
• I don't think it is any value in (-1,1) as proven in my main post.The sample median in my main post is either -1 or 1. But the asymptotic median is certainly 0. – NamelessGods Oct 6 '20 at 20:37