Real Applications of Markov's Inequality When is Markov's Inequality useful?  It seems to me that it's a very rough upper bound.  For example, if we roll a die and want to know the probability of the result being a 5 or greater we have that there is at most a 3.5/5 chance.  That's huge relative to the actual chance.  Am I misunderstanding?
 A: Although Markov's inequality is a rough bound, it has plenty of applications. For example, if we are interested in the tail behavior of
$$\mathbb{P}(|X| \geq R)$$
(i.e. the asymptotic behavior for large $R$), then by Markov's inequality,
$$\mathbb{P}(|X| \geq R) \leq \frac{1}{R^p} \mathbb{E}(|X|^p)$$
whenever $\mathbb{E}(|X|^p)<\infty$. This means that $\mathbb{P}(|X| \geq R)$ decays (at least) as $\frac{1}{R^p}$.

Example: Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of random variables such that $$\mathbb{E}(X_n^2) \leq \frac{1}{n^2}.$$ Then $$\lim_{n \to \infty} X_n = 0 \qquad \text{almost surely}.$$

Proof: By the Borel-Cantelli lemma it suffices to show $$\sum_{n=1}^{\infty} \mathbb{P}(X_n>\epsilon)<\infty$$ for any $\epsilon>0$. This follows directly from Markov's inequality: $$\sum_{n=1}^{\infty} \mathbb{P}(X_n>\epsilon) \leq \frac{1}{\epsilon^2} \sum_{n=1}^{\infty} \mathbb{E}(X_n^2) \leq \frac{1}{\epsilon^2} \sum_{n=1}^{\infty} \frac{1}{n^2}.$$

Example: Suppose that $X_n \to X$ in $L^1$. Show that $X_n \to X$ in probability.

Proof: This is a direct consequence of Markov's inequality.
A: Markov's Inequality and its corollary Chebyshev's Inequality are extremely important in a wide variety of theoretical proofs, especially limit theorems.
A previous answer provides an example.
In practice, these inequalities have sometimes been used to find bounds
on probabilities arising in practice, hoping that the bounds would approximate
the probabilities of interest. This was especially true when the probabilities
involved distributions for which printed tables were not available, or where
the distributions were nonstandard or complicated. Usually, this is not a
very good method of approximation for practical purposes.
Nowadays, software packages permit exact computations for a wide variety of frequently used distributions and allow approximation by simulation in nonstandard and complicated cases.
Example 1: Let $X \sim$ Gamma(shape=5, rate=0.1). Then $E(X) = 50$ and
Markov's Inequality gives $P(X \ge 100)\le 50/100 = 1/2,$ whereas a statistical computer package gives $P(X > 100) = 0.0293.$ The bound is certainly true,
but hardly of practical use.
Example 2: Suppose the length of time to complete a process is the sum of three
known independent distributions: (1) $W \sim$ UNIF(0,12), with $E(W) = 6,$ and $V(W) = 12$; (2) $X \sim$ EXP(rate = 1/5),  $E(X) = 54$ and  $V(X) = 25$; and (3) $Y \sim$ NORM(10, sd = 2), with $E(Y) = 10$ and $V(Y) = 4.$
Then $T = V + X + Y$ has a distribution for which the density function
and CDF are very difficult to find, but we know that $E(T) = 21$ and $V(T) = 41.$ We seek $P(T > 35)$, the probability that the overall process takes more than 35 time units to completion. We mention several possible methods to bound or approximate this probability.
Markov:  Because of the normal component, there is a tiny probability that $T$ is negative, and so strictly speaking Markov's Inequality does not apply apply to $T.$ But the positive part $T^+$ of $T$ is essentially the same as $T$, and we ignore this difficulty.
Then Markov's Inequality says $P(T > 35) \leq 21/35 = 0.6$.
Chebyshev: From Chebyshev's inequality we know that
 $P(|T-21| \ge 6.403k) \le 1/k^2.$
Taking $k = 2.1875,$ and ignoring negative values, 
we have $P(T \ge 35) \le 1/k^2 = 0.21.$
Normal approximation: Because $T$ is the sum of three 
independent (even though very different) random variables, we might
venture to try a normal approximation, which gives $P(T > 35) \approx 0.0144.$
Simulation. The brief R program below does the simulation.
 m = 10^6
 w = runif(m, 0, 12);  x = rexp(m, 1/5);  y = rnorm(m, 10, 2)
 t = w + x + y
 mean(t);  var(t);  mean(t > 35)

Results are consistent with known values $E(T) = 21$ and $V(T) = 41$
The last statement gives $P(T > 35) = 0.03,$ very likely accurate to two places. A histogram (not
reproduced here) shows that the simulated distribution of $T$ is
moderately right skewed (inherited from its exponential component),
which explains why the normal approximation "missed" some of this
right-tailed probability.
Closing comment: Without taking anything away from the well established
wide applicability of Markov's Inequality to prove theoretical
results, I have to say that I have not seen an example in which
it or related inequality gives a serviceable numerical approximation in an
applied setting. If anyone knows of a favorable example, I would be very happy to see it.
