# Convergence of Uniform Distribution

How do you prove that the sample median of $X_i, \ldots, X_n$ where $i$ is odd converges to the median of U$(0,1)$? Do you have to find the cdf of the median, find the p.m.f. of median, and then show that as $n$ increases, the variance of the p.m.f. gets smaller, eventually becoming a point mass around the true median?

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Since your post is incomplete, I will answer the question in my interpretation. Assume $X_i$ are independent identically distributed standard uniform random variables. We seek to show that the median of a sample of size $n = 2m+1$ converges in probability to $\frac{1}{2}$ as $m \to \infty$.

See wikipedia article on order statistics, and for the expression for the probability density of the median $X_{(m+1)}$, which is $m+1$-st smallest element in the sample. $$f_{X_{(m+1)}}(u)\mathrm{d} u = \frac{(2m+1)!}{(m)!^2} u^{m} (1-u)^{m} \mathbf{1}_{0 < u < 1} \mathrm{d} u = (2m+1)\binom{2m}{m} (u(1-u))^{m} \mathbf{1}_{0 < u < 1} \mathrm{d} u$$

By definition of convergence in probability, we seek to evaluate: $$\begin{eqnarray} p_{m,\epsilon} &=&\mathbb{P}\left( \vert X_{(m+1)} - \frac{1}{2} \vert \ge \epsilon \right) = 2 \int_0^{\frac{1}{2}-\epsilon} \frac{(2m+1)!}{(m)!^2} u^{m} (1-u)^{m} \mathbf{1}_{0 < u < 1} \mathrm{d} u \\ &=& 2 \int_0^{\frac{1}{2}-\epsilon} (2m+1)\binom{2m}{m} (u(1-u))^{m} \mathrm{d} u \\ & \le & (2m+1)\binom{2m}{m} \left( \frac{1}{4} - \epsilon^2 \right)^m \int_0^{\frac{1}{2} -\epsilon} 2 \mathrm{d} u \end{eqnarray}$$ The last inequality follows because $u(1-u)$ is increasing for $0<u<\frac{1}{2}- \epsilon$.

We should prove that for any fixed $\epsilon >0$, $\lim_{m \to \infty} p_{m, \epsilon} = 0$.

For large $m$, $\binom{2m}{m} \sim \frac{4^m}{\sqrt{m \pi}}$, therefore $$\begin{eqnarray} 0 \le \lim_{m \to \infty} p_{m, \epsilon} &\le& \lim_{m \to \infty} (2m+1)\binom{2m}{m} \left( \frac{1}{4} - \epsilon^2 \right)^m (1-2 \epsilon) \\ &=& \lim_{m \to \infty} (2m+1) \frac{4^m}{\sqrt{m \pi}} \left( \frac{1}{4} - \epsilon^2 \right)^m (1-2 \epsilon) \\ &=& \lim_{m \to \infty} \frac{2m+1}{\sqrt{m \pi}} \left( 1 - 4 \epsilon^2 \right)^m (1-2 \epsilon)= 0 \end{eqnarray}$$

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If you were to shift the distribution to U(theta+ 0.5,theta-0.5) how would the distribution change? – lord12 Nov 17 '11 at 16:45
@lord12 You know the density of $X$, now you should find the density of $X + \theta - \frac{1}{2}$. Do you know how ? – Sasha Nov 17 '11 at 16:47
Do you just change the lower bound for the integration to theta - 1/2? You want P(|X-theta} > e)= 0 since theta is median of this distribution. – lord12 Nov 17 '11 at 17:29
In the integration, I don't get how you just substituted in (1/4-e^2) without integrating u(1-u)^m – lord12 Nov 17 '11 at 17:59
@lord12 Since $u(1-u)$ is increasing on the domain $0<u< v = \frac{1}{2} -\epsilon$, it follows that $\int_0^v (u(1-u))^m du \le \int_0^v (v(1-v))^m d u = (v(1-v))^m v$. – Sasha Nov 17 '11 at 19:57