Poisson Distribution and Markov's Inequality

"The random variable X has a Poisson distribution with parameter λ = 3. By employing Markov’s inequality,

$\displaystyle P(|X|\geq\epsilon)\leq\frac{E(|X|k)}{\epsilon^k}$

calculate bounds for P (X ≥ 7) in the cases where k = 1, 2. Which of the two estimates is better?"

What I've done:

$\displaystyle Using\ P_X(k)=\frac{e^{-\lambda}\lambda^k}{k!}$

$\displaystyle with\ \lambda=3\ and\ k=1:\ P_X(k)=0.14936$

$\displaystyle with\ \lambda=3\ and\ k=2:$$\displaystyle \ P_X(k)=0.22404$

I'm not sure if thats right or have any idea which is a better estimate. Could someone help explain please? Thanks

Re: Poisson Distribution and Markov's Inequality

Quote:

Originally Posted by

**carla1985** "The random variable X has a Poisson distribution with parameter λ = 3. By employing Markov’s inequality,

$\displaystyle P(|X|\geq\epsilon)\leq\frac{E(|X|k)}{\epsilon^k}$

calculate bounds for P (X ≥ 7) in the cases where k = 1, 2. Which of the two estimates is better?"

What I've done:

$\displaystyle Using\ P_X(k)=\frac{e^{-\lambda}\lambda^k}{k!}$

$\displaystyle with\ \lambda=3\ and\ k=1:\ P_X(k)=0.14936$

$\displaystyle with\ \lambda=3\ and\ k=2:$$\displaystyle \ P_X(k)=0.22404$

I'm not sure if thats right or have any idea which is a better estimate. Could someone help explain please? Thanks

Hi carla1985! :)

You seem to be mixing up the symbol k.

Let's disambiguate by *not *using k.

Let's write Markov's inequality as $\displaystyle P(X\geq a)\leq\frac{E(X^m)}{a^m}$.We'll look at cases where m=1,2.

Furthermore, note that a Poisson distribution has $\displaystyle EX=\lambda$ and $\displaystyle \sigma^2=\lambda$.

Then from the PDF of the Poisson distribution, we know that: $\displaystyle P(X \ge 7) \ge P(X=7) = \frac{e^{-\lambda}\lambda^7}{7!}$.

And from Markov's inequality with m=1 we get $\displaystyle P(X\geq 7)\leq\frac{E(X^1)}{7^1} = \frac{\lambda}{7}$.

Re: Poisson Distribution and Markov's Inequality

Ah ok, so with m=1 $\displaystyle \frac{\lambda}{7}$

so with m=2 $\displaystyle EX=\lambda+\lambda^2$

so $\displaystyle \frac{\lambda+\lambda^2}{7^2}=\frac{12}{49}$?

which would then make m=2 the better estimate :)

Re: Poisson Distribution and Markov's Inequality

Quote:

Originally Posted by

**carla1985** Ah ok, so with m=1 $\displaystyle \frac{\lambda}{7}$

so with m=2 $\displaystyle EX=\lambda+\lambda^2$

so $\displaystyle \frac{\lambda+\lambda^2}{7^2}=\frac{12}{49}$?

which would then make m=2 the better estimate

I think you meant $\displaystyle E(X^2)=\lambda+\lambda^2$.

But yes! :cool:

Re: Poisson Distribution and Markov's Inequality

yes thats what i meant. thank you very much :D