Let $X\sim \mathrm{Poisson}(\lambda)$, show that $P(X\le \lambda/2) \le 4/\lambda$

Question

Let $X\sim\mathrm{Poisson}(\lambda)$, show that $P(X\le \lambda/2) \le 4/\lambda$.

It is clear for $\lambda \in ]0,4]$ because in this case $4/\lambda \ge 1$. But for other values of $\lambda$ I do not see the point.

Answer 1

The mean and the variance of the Poisson distribution of parameter $\lambda$ both are $\lambda$.

So,

$$P\left(X\le\frac{\lambda}2\right)=P\left(X-\lambda\le -\frac{\lambda}2\right).$$ Squaring both sides of the inequality, we get

$$P\left(X\le\frac{\lambda}2\right)=P\left((X-\lambda)^2 \le\frac{\lambda^2}4\right).$$

Then by Markov's inequality

$$P\left(X\le\frac{\lambda}2\right)\le\frac4{\lambda^2}E[(X-\lambda)^2]$$

Since $E[(X-\lambda)^2]=\lambda$, the variance, finally we get what was to be proved.

Yes, this is Chebyshew's inequality.

Let's see, how to prove Markov's inequality for a non negative discrete random variable. The expectation is

$$E[X]=\sum_{i=1}^{\infty}x_ip_i\ge a\cdot\sum_{\{j: x_j\ge a\}}p_j=aP(X\ge a),$$

for any non negative $a$. Hence, $$P(X\ge a)\le \frac{E[X]}a.$$