# Using Chebyshev's inequality, show $P(X \geq 2\lambda) \leq \frac{1}{\lambda}$

Let $$X \sim \mathsf{Poisson}(\lambda),$$ show $$P(X \ge 2\lambda) \le \frac{1}{\lambda}$$.

I am not understanding this fully, Chebyshev's inequality has the absolute value, what I did was start off with:

$$P(|X - \lambda| \geq 2\lambda) \leq \frac{ \lambda}{ 4\lambda^{2}}$$ but I'm unsure where to go from here. Do I need to expand the inside and remove the absolute value sign?

• I haven’t written out the whole proof, but I think it will help you to notice that $\sigma$ for a Poisson distribution is $\sqrt{\lambda}$. I know you know this, but it might help to see that made explicit. (Now I just wrote it out. My hint should help.) – Dave Oct 4 '20 at 0:08
• Start by manipulating the $P(X\ge 2\lambda)$ you’re given. – Dave Oct 4 '20 at 0:15
• I don't see how the standard deviation really helps me mate. Yes I have manipulated it, but I'm unsure, do I need to change the $P(X \geq 2\lambda)$ into something where the inside has a absolute value? – Wallace Oct 4 '20 at 0:17
• You’ll need an absolute value in there, yes, but that was my second step, not my first. To do my first step, think about what has to be in the absolute value. – Dave Oct 4 '20 at 0:19

Hint: $$P(X \ge 2\lambda) = P(X - \lambda \ge \lambda) \le P(|X-\lambda| \ge \lambda)$$ (To justify the inequality, note that $$X-\lambda \ge \lambda$$ implies $$|X-\lambda| \ge \lambda$$.) Then apply Chebychev's inequality to the right-hand side.