Let X have $f(x) = \begin{cases} 1, & \text{if $0<x<1/2$} \\[2ex] 2, & \text{if $3/4<x<1$} \\[2ex] 0, & \text{othewise} \end{cases}$
Find all the points that will qualify as being median of X: I know that $$F_X(x)=\int_0^{1/2}(1)dy+\int_{1/2}^{3/4}(0)dy+\int_{3/4}^{1}(2)dy+\int_{1}^{x}(0)dy$$ was the answer laid out by my lab instructor although this sounds clear I cannot see how to fully draw the CDF from this expression. Since I know that when I draw it I'll just have to select the interval that contains 1/2 as a probability.
I would agree that $F_X(M_X)=1/2$ with $M_X$ being the Median point. And I would agree that: $$F_X(x)=\int_0^{x}(1)dy+\int_{1/2}^{x}(0)dy+\int_{3/4}^{x}(2)dy+\int_{1}^{x}(0)dy$$
I would like to get some clarification on that subject.
Sorry for the poorly formed question...
I very much appreciate your input. But I think I may be lacking a very deep understanding of these expressions.
I know that $F_X(x)=P(X\le x)= \int _ 0^xf(t)dt$. I did try to integrate the whole thing and, as expected it gave me 1.
I can understand most of what you're saying but I don't know why you have some x on some of your integrals and not some.
I haven't be shown that way of writing this expression. PLus I drew $f$ and I know f(x)= 0 between 1/2 and 3/4 which means F(x) on that interval gives a constant but since on the pdf(or the question) I don't have a value to integrate I'm lost. I figured doing $\int_{1/2}^{3/4}f(t)dt$
Let $F(x) = \int_{-\infty}^x f(t) dt$. Then just integrating gives $F(x) = \begin{cases} 0,& x \le 0 \\ x, & x \in [0, {1 \over 2}] \\ {1 \over 2}, & x \in [{1 \over 2}, {3 \over 4}] \\ {1 \over 2} + 2(x-{3 \over 4}), & x \in [ {3 \over 4}, 1] \\ 1, & x \ge 1 \end{cases}$
Aside: We can write $F(x) = \int_{-\infty}^x (1 \cdot 1_{(0,{1 \over 2})}(t) + 2 \cdot 1_{({3 \over 4}, 1)} (t)) dt $.
Look for values $m$ such that $P[ X \ge m] \ge {1 \over 2}$ and $P[ X \le m] \ge {1 \over 2} $.
The values of $m$ that satisfy $P[ X \le m] \ge {1 \over 2} $ are $[{1 \over 2}, \infty)$ and the values of $m$ that satisfy $P[ X \ge m] \ge {1 \over 2}$ are $(-\infty, {3 \over 4}]$.