The annual return, R, of a certain stock is a random variable with mean 10. Use Markov’s inequality to obtain a bound for the probability of the stock return being at least 20. Assuming now that R has an exponential distribution with probability density function
f(r) = (1/10)e^(-r/10) r>0, 0 Otherwise
calculate the true value of P(R ≥ 20) and compare this with your bound.
I have the inequality equalling .5 and I think I have the density function but I am stuck on calculating P(R ≥ 20)
Using Markov's inequality, we have
$$ \mathbb{P}\left(R\ge 20\right)\le \frac{\mathbb{E}(R)}{20}=\frac{10}{20}=0.5$$
From the assumption, $R\sim Exp(1/10)$. In general, if $X$ is an exponential random variable with parameter $1/\lambda$, the its probability density function and cummulative distribution function are
$$f(x) = \frac{1}{\lambda}e^{-x/\lambda},\quad x>0$$
and
$$F(x) = 1 - e^{-x/\lambda},\quad x>0$$
Hence, the cummulative distribution function of $R$ is
$$F(r) = \mathbb{P}(R < r) = 1 - e^{-r/10}$$
Then,
$$\mathbb{P}(R\ge 20)= 1- \mathbb{P}(R<20)=1-F(20)= 1-(1-e^{-2})=e^{-2}=0.135 $$
And you can see clearly that $\mathbb{P}(R\ge 20) < 0.5$ satisfying Markov's inequality.