Suppose $Y|\Lambda\sim U(0,\lambda)$ with $\Lambda \sim U(0,1)$. If there is sample with size $n$ of $Y$ (To simplify, assume $n$ is odd, so $n=2m-1$). How do I calculate the expected value of median ($E[Y_{(m)}]$) and variance of the median ($Var[Y_{(m)}]$). Thanks in advance

Note: To clarify, my intention was that $\{ Y_1, \dots, Y_n \}$ is generated with the same $\Lambda$ value (so there's only 1 value of $\Lambda$, not $n$ for each value of $Y_i$). Sorry for the lack of information on the previous post.

  • 4
    $\begingroup$ You can (easily) find the distribution of $Y,$ which directly gives the distribution of the median, and then apply the standard integral formulas, giving a three-step solution. Which step do you need help with? $\endgroup$
    – whuber
    Sep 15, 2023 at 18:04
  • $\begingroup$ Why should it not converge? It is bounded between $0$ and $1$ and there is no reason to suppose it would keep changing direction as $m$ keeps increasing $\endgroup$
    – Henry
    Sep 15, 2023 at 19:27
  • $\begingroup$ Re-stated: Finding the mixture distribution of 𝑌 is okey-dokey. Finding the distribution of the median is okey-dokey too. But, it seems to me the expectation of the median does not have a closed-form. Are you saying that it does? $\endgroup$
    – wolfies
    Sep 15, 2023 at 19:33
  • 2
    $\begingroup$ @Wolfies By symmetry, the expectation of the median conditional on $\Lambda$ is $\Lambda/2.$ Thus, the expectation of the median of $Y$ must be $1/4.$ // The exact distribution of the median is straightforward to find, because it's a uniform scale mixture of Beta distributions. $\endgroup$
    – whuber
    Sep 15, 2023 at 19:57
  • 1
    $\begingroup$ @whuber: I agree with you, but to clarify the difference between the two readings of the questions try with $n=9$ in the R code medA<-function(n){median(runif(n,0,runif(1)))}; simsA<-replicate(10^4,medA(9)); mean(simsA) which gives about $0.25$ while medB<-function(n){median(runif(n,0,runif(n)))}; simsB<-replicate(10^4,medB(9)); mean(simsB) gives about $0.20$. Wolfies has $n$ different $\Lambda$ in each sample, while you and I and Zhanxiong have the same value through that sample $\endgroup$
    – Henry
    Sep 16, 2023 at 23:15

3 Answers 3


Caveat: The calculation below relies on the (not explicitly stated) condition that the sample $\{Y_1, \ldots, Y_n\}$ is drawn independently from $U(0, \lambda)$ once $\Lambda = \lambda$ was observed. In other words, $Y_1, \ldots, Y_n$ are conditionally independent given $\Lambda = \lambda$. Note that this sampling mechanism is essentially different from drawing $n$ i.i.d. observations from the marginal density (as simulated in @wolfies answer) $f_Y(y) = -\log y, 0 < y < 1$. The closed-form solutions with the second interpretation should be much harder -- the density of $Y_{(m)}$ if $Y_1, \ldots, Y_{2m - 1} \text{ i.i.d. } \sim f_Y$ is \begin{align} g(x) = -\frac{(2m - 1)!}{[(m - 1)!]^2}[x(1 - \log x)(1 - x(1 - \log x))]^{m - 1}\log x, \quad 0 < x < 1. \end{align}

Here we use that the $k$-th order statistic of the standard uniform distribution is beta-distributed, i.e., $U_{(k)} \sim B(k, n + 1 - k)$, whence $\lambda^{-1}Y_{(m)}|\Lambda = \lambda \sim B(m, 2m - 1 + 1 - m) = B(m, m)$. It then follows by the moments of beta distribution that \begin{align} & E[\lambda^{-1}Y_{(m)}|\Lambda = \lambda] = \frac{m}{m + m} = \frac{1}{2}, \\ & \operatorname{Var}(\lambda^{-1}Y_{(m)}|\Lambda = \lambda) = \frac{m^2}{4m^2\cdot(2m + 1)} = \frac{1}{8m + 4}. \end{align} Therefore \begin{align} E[Y_{(m)}|\Lambda] = \frac{1}{2}\Lambda, \quad \operatorname{Var}(Y_{(m)}|\Lambda) = \frac{\Lambda^2}{8m + 4}. \end{align} It then follows by the law of total expectations and variances that \begin{align} & E[Y_{(m)}] = \frac{1}{2}E[\Lambda] = \frac{1}{4}, \\ & \operatorname{Var}(Y_{(m)}) = E[\operatorname{Var}(Y_{(m)}|\Lambda)] + \operatorname{Var}(E[Y_{(m)}|\Lambda]) = \frac{1}{8m + 4}E[\Lambda^2] + \frac{1}{4}\operatorname{Var}(\Lambda) \\ & = \frac{1}{8m + 4}\times\frac{1}{3} + \frac{1}{4}\times \frac{1}{12} = \frac{2m + 5}{96m + 48}. \end{align}

  • $\begingroup$ @wolfies of course not... Did you really understand the argument? Take $n = 2m - 1$ and $k = m$ in the first link in my answer. Here $n$ can be any positive integer. $\endgroup$
    – Zhanxiong
    Sep 16, 2023 at 13:45
  • $\begingroup$ At its essence, your solution posits that the expected value of the sample median is a constant (independent of the sample size). I have posted a solution that asserts that it is not a constant. It is very easy to check via simulation whether it is a constant, or not. $\endgroup$
    – wolfies
    Sep 16, 2023 at 15:42
  • $\begingroup$ @wolfies I think the difference of your simulation and my calculation above lies in that how the sample $Y$ was drawn (the OP didn't sufficiently clarify it) -- my interpretation is that a sample was drawn given the same $\Lambda = \lambda$, that is $Y_1, \ldots, Y_n$ are (conditionally) independent $U(0, \lambda)$ given $\Lambda = \lambda$, while your simulation is based on the interpretation that $Y_1, \ldots, Y_n \text{ i.i.d. } \sim f(y) = -\log y$. I will clarify it in my answer. $\endgroup$
    – Zhanxiong
    Sep 16, 2023 at 16:50
  • $\begingroup$ It occurred to me too that this explains the difference. However, we both agreed that the unconditional pdf of $Y$ is $-\log y$, and that result is not consistent with holding $\lambda$ constant. $\endgroup$
    – wolfies
    Sep 16, 2023 at 17:12
  • $\begingroup$ @wolfies Again, that depends on how you interpret the problem. $-\log y$ is the unconditional pdf for ONE $Y$, but not the the joint unconditional pdf of $(Y_1, \ldots, Y_n)$ if we take the first interpretation, under which case the expectation is indeed $1/4$. $\endgroup$
    – Zhanxiong
    Sep 16, 2023 at 17:17

As whuber has noted, this problem has 3 stages:

Step 1: The unconditional pdf of $Y$

The first stage is comparatively easy (see here for instance) which is to find the unconditional distribution of $Y$, i.e. the parameter mixture of two Uniforms.

The unconditional pdf of $Y$, say $f(y)$ is :

enter image description here

Step 2: The distribution of the sample median of $Y$

The OP assumes that a random sample of size $n$ is drawn on $Y$ where $n$ is odd, and so we write $n=2r-1$. The sample median, which we denote by $M$, corresponds to the middle order statistic in a sample of size $n$, which is by definition thus $Y_{(r)}$.

The pdf of the $r^\text{th}$ order statistic, in a sample of size $n$, can be found using the OrderStat[r, f, n] function in the mathStatica package for Mathematica.

The pdf of the sample median is then $g(m)$:

enter image description here

with domain of support:

enter image description here

Step 3: The mean and variance of the sample median

Calculating the mean and variance of the sample median does not appear to have an obvious closed form solution, as a general function of the sample size ($n$ or equivalently $r$). However, for given values of $r$, it yields exact solutions. For instance, when $r = 1, 2, 3, 4, 5$:

enter image description here

Note that this is the mean of the sample median for sample sizes $n = 1, 3, 5, 7, 9$, and corresponds to numerical values of:

{0.25, 0.220486, 0.209634, 0.204037, 0.200629}


Similarly, the variance of the sample median, when $r = 1, 2, 3, 4, 5$ is:

enter image description here

Note that this is the variance of the sample median for sample sizes of $n = 1, 3, 5, 7, 9$, and corresponds to numerical values of:

{0.0486111, 0.0242859, 0.0159245, 0.0117921, 0.00934636}

The following diagram plots the expectation of the sample median, for $r$ from 1 to 25 (corresponding to the odd-sized samples from $n = 1$ to $49$):

enter image description here

Check by Simulation

Here is some quick code to simulate: for any given value of $r$, generate 500,000 samples of size $n = 2r-1$. Then, for each sample, calculate the sample median, and then find the sample mean of all 500,000 sample medians:

SimExpectedMedian[r_] := (data = Partition[RandomNumber[(2 r - 1)*500000, f], 2 r - 1]; Mean[Map[Median, data]])

For example, here we simulate the solution for $r = 1, 2, 3, 4$ and $5$:

enter image description here

... which very closely fits the exact solutions derived above (stated here numerically for ease of comparison):

{0.25, 0.220486, 0.209634, 0.204037, 0.200629}

The following diagram plots:

a) the EXACT expectation of the sample median derived above (for $r = 1$ to 25): big BLUE dots

b) the SIMULATED mean of the sample median: smaller RED dots

enter image description here

The simulated and exact results are so close that red and blue dots lie on top of each other.

  • $\begingroup$ I read the original question as saying $\Lambda$ is the same value for all $n$ samples of $Y$. Repeat the experiment and $\Lambda$ will change. You seem to read it as having $n$ different values of $\Lambda$ in the sample of $Y$. $\endgroup$
    – Henry
    Sep 16, 2023 at 23:20

To contribute to this thread that has already seen very interesting answers:

A. The marginal distribution function of $Y$ is

$$F_Y(y) = y\cdot(1-\ln(y)).$$

To find the population median, we have $$m_Y:= F_Y(m) = \frac 12 \implies m_Y\cdot(1-\ln(m_Y)) = \frac 12 \implies m_Y: 1-\ln(m_Y)-\frac 1 {2m_Y} = 0.$$ This implicit equation is monotonic and it is satisfied for $$m_Y \approx 0.18668235.$$

This is the median of the marginal distribution of $Y$, and we see that the expected value of the sample median from draws from this marginal distribution (@wolfies answer), appears to tend downwards from the value $0.25$ as the sample size increases, so we can expect it to become unbiased at the limit.

B. The conditional distribution function of $Y$, evaluated at the conditional median $m^c_Y$ is

$$\Pr\big(Y \leq m^c_Y \mid \Lambda = \lambda\big) = \frac {m^c_Y}{\lambda} = \frac 12 \\ \implies m^c_Y = \frac {\lambda} 2.$$

Viewed as a function of $\Lambda$, this is a random variable, and so, the expected value of the conditional median is, (with $f_{\Lambda}(\lambda)$ the density of $\Lambda$)

$$E[m^c_Y] = \int_0^1 f_{\Lambda}(\lambda)\cdot m^c_Y d\lambda = \int_0^1 1\cdot \frac {\lambda} 2\,d\lambda = \frac {\lambda^2} 4 \Big |^1_0 = \frac 1 4.$$

We see that

  • the "expected value of the conditional median"


  • the "unconditional expected value of the sample median drawn from the conditional distribution" (@Zhanxiong answer).

This tells us that the sample conditional median is unbiased for the population conditional median, already for finite samples.


Why the difference in performance? Why, when we draw from the marginal distribution of $Y$, we get unbiasedness of the sample median only asymptotically, while when we draw from a conditional (without needing to know the value of $\lambda$, just that it has been fixed), we get unbiasedness immediately? Contemplate.


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