medianprobabilityself-study
By definition of median, i.e.
$P(X\leq m)\geq 1/2 \text{ and } P(X\geq m)\geq 1/2.$
What is the median of Bernoulli distribution with a probability parameter of $p=0.2$ ($P(X=1)=0.2$)?
Suppose $m$ is the median. Then $P(X\leq m)\geq 1/2$ implies $m\geq0$. $P(X\geq m)\geq 1/2$ implies $m<0$. One says m is greater or equal to 0 and the other says it is strictly less than 0. I don't understand where I did wrong.
Here are the CDF plot
.
Here is a hint.
Consider carefully the term $\mathbb P( X \leq z Y \mid z Y < 1 )$. In particular, for concreteness, choose $z = 2$, so that we are considering the event $\mathbb P( X \leq 2 Y \mid Y < 1/2 )$.
Now, look at this picture (which is very closely related to the above probability).
Now, does that conditional probability depend on our particular choice of $z$?
We wish to show that $\hat \theta_n \xrightarrow{P} \theta_0$. By definition, this states that given $\epsilon > 0$ $$ P(|\hat \theta_n - \theta_0| > \epsilon) \rightarrow 0 \text{ as } n \rightarrow \infty. $$
Given $X_1,..., X_n$ are iid with finite $n$, we see that if we were to reorder $\{X_i\}$ in ascending order, the definition of $\hat \theta_n$ gives $\hat \theta_n = X_{\lceil n/2 \rceil}$. From this we have that $$ \hat F_n(\hat \theta_n) = \left \{ \begin{array}{ll} 1/2 & \text{ if n is even} \\ \frac{1}{2} + \frac{1}{2n} & \text{ if n is odd.} \end{array} \right. $$
Using this, we can calculate \begin{align} P(|\hat \theta_n - \theta_0| > \epsilon) &\leq P(|\hat \theta_n - \theta_0| \geq \epsilon) \\ &= P(\hat \theta_n \leq \theta_0 - \epsilon) + P(\theta_n + \epsilon \leq \hat \theta_n) \\ &= P(\hat F_n(\hat \theta_n) \leq \hat F_n(\theta_0 - \epsilon)) + P(\hat F_n(\theta_n + \epsilon) \leq \hat F_n(\hat \theta_n)) \end{align} where the last equality is due to the fact that $\hat F_n(x)$ is non-decreasing. Notice that because $\hat F_n(x)$ is non-decreasing and not strictly increasing, it can only preserve weak inequalities.
We continue by analyzing the terms individually. First, consider the term $$ P(\hat F_n(\hat \theta_n) \leq \hat F_n(\theta_0 - \epsilon)). $$ When $n$ is even, \begin{align*} P(1/2 \leq \hat F_n(\theta_0 - \epsilon)) &= P(1/2 - F(\theta_0 - \epsilon)) \leq \hat F_n(\theta_0 - \epsilon) - F(\theta_0 - \epsilon))\\ &\rightarrow 0 \text{ as } n \rightarrow \infty \end{align*} where this last fact follows from the fact that $1/2 - F(\theta_0 - \epsilon) > 0 $ (we can see this from the defintion of the median) and that $\hat F_n(x) \xrightarrow{p} F(x)$, as we demonstrated previously. Similarly, when $n$ is odd, \begin{align*} P(1/2 + \frac{1}{2n} \leq \hat F_n(\theta_0 - \epsilon)) &= P(1/2 + \frac{1}{2n} - F(\theta_0 - \epsilon)) \leq \hat F_n(\theta_0 - \epsilon) - F(\theta_0 - \epsilon))\\ &\rightarrow 0 \text{ as } n \rightarrow \infty. \end{align*} Now consider the term $$ P(\hat F_n(\theta_n + \epsilon) \leq \hat F_n(\hat \theta_n)). $$ Similar to before, we have \begin{align*} P(\hat F_n(\theta_n + \epsilon) \leq \hat F_n(\hat \theta_n)) &= P(-\hat F_n(\theta_n + \epsilon) \geq -\hat F_n(\hat \theta_n)) \\ &= P(F(\theta_n + \epsilon) -\hat F_n(\theta_n + \epsilon) \geq F(\theta_n + \epsilon) -\hat F_n(\hat \theta_n)) \\ &\rightarrow 0 \text{ as } n \rightarrow \infty. \end{align*} This last fact comes again from the fact that that $\hat F_n(x) \xrightarrow{p} F(x)$ and that $F(\theta_n + \epsilon) -\hat F_n(\hat \theta_n) > 0$, which we have because the uniqueness of the median gives that for $\epsilon > 0$ we have $F(\theta_0 + \epsilon) > .5$.
These two results, applied back give \begin{align*} P(|\hat \theta_n - \theta_0| > \epsilon) &\leq P(|\hat \theta_n - \theta_0| \geq \epsilon) \nonumber \\ &\rightarrow 0 \text{ as } n \rightarrow \infty \end{align*} and thus $$ \hat \theta_n \xrightarrow{p} \theta_0. $$
Best Answer
Let $X \sim \mathsf{Bern}(p=.2)\equiv\mathsf{Binom}(n=1, p=.2).$ In R, where
qbinomis the inverse CDF (quantile function) of a binomial distribution a median $\eta = 0.$$P(X \le 0) = P(X = 0) = 0.8 \ge 1/2.$
And obviously, $P(X \ge 0) = 1 \ge 1/2.$
The CDF of $X$ is plotted below. The median of $X$ is taken to be the value at which the CDF 'curve' is (or 'crosses') $1/2.$
Also, for context, if we simulate $1000$ observations from this distribution, we get $805$ Failures (0) and $195$ Successes. According to R, the sample median is also $0.$