Working on this question:
The median of a continuous random variable with CDF $F(x)$ is the value $m$ that guarantees that $P\{X > m\} = P\{X < m\} = \frac{1}{2}$ The mode is the value of X where the PDF attains its maximum. What are the median and the mode of:
a) the uniform random variable on interval [0,1];
b) the exponential random variable with rate $\lambda$?
Attempt at a solution:
a) median-
Pr{$X < m$} = $1/2$
$\int(P.D.F)$ = 1/2 from lower bounds of x to $m$ (where $m$ is median)
in this case, since the R.V. has a uniform distribution, we can assume that the graph is a rectangle...
However, I am having trouble coming up with the P.D.F of this R.V... I know that a uniform distribution means that the
b) mode
Since this is an exponential R.V. with rate $\lambda$, we are looking for the maximum point on it's graph.
At this point, $f'(x) = 0$, so we would want to differentiate the P.D.F, set that equal to zero, and solve for $x$ (this would make $x$ our mode).
I'm getting confused by the lack of P.D.F function... how should I approach the solution to this problem, for parts A and B?