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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?

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up vote 2 down vote accepted

For a random variable uniform on $[0,1]$ the density function is $1$ on $[0,1]$ and $0$ elsewhere. Median should be easy to pick up. The density function attains a maximum everywhere on our interval. Lotsa modes!

Let $X$ have exponential distribution parameter $\lambda$. Then $X$ has density function $f_X(x)=\lambda e^{-\lambda x}$ for $x\ge 0$, and $0$ elsewhere. This is a steadily decreasing function (if you wish, take the derivative, but that should not be necessary). You can simply look at the graph.

So the mode is at $x=0$. For the median $m$, we want $P(X<m)=P(X>m)=1/2$. Let's find $P(X<m)$. We have $$P(X<m)=\int_0^m \lambda e^{-\lambda x}\,dx.$$ An antiderivative is $-e^{-\lambda x}$. So the definite integral is $1-e^{-\lambda m}$. Set this equal to $1/2$. We want $e^{-\lambda m}=1/2$. I prefer to write this as $e^{\lambda m}=2$. To find $m$, take the natural logarithm of both sides.

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