Poisson Distribution and Markov's Inequality

• Apr 14th 2013, 02:54 AM
carla1985
Poisson Distribution and Markov's Inequality
"The random variable X has a Poisson distribution with parameter λ = 3. By employing Markov’s inequality,

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

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

What I've done:

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

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

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

• Apr 14th 2013, 07:58 AM
ILikeSerena
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,

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

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

What I've done:

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

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

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

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 $P(X\geq a)\leq\frac{E(X^m)}{a^m}$.We'll look at cases where m=1,2.

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

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

And from Markov's inequality with m=1 we get $P(X\geq 7)\leq\frac{E(X^1)}{7^1} = \frac{\lambda}{7}$.
• Apr 14th 2013, 08:49 AM
carla1985
Re: Poisson Distribution and Markov's Inequality
Ah ok, so with m=1 $\frac{\lambda}{7}$

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

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

which would then make m=2 the better estimate :)
• Apr 14th 2013, 08:58 AM
ILikeSerena
Re: Poisson Distribution and Markov's Inequality
Quote:

Originally Posted by carla1985
Ah ok, so with m=1 $\frac{\lambda}{7}$

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

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

which would then make m=2 the better estimate

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

But yes! :cool:
• Apr 14th 2013, 08:59 AM
carla1985
Re: Poisson Distribution and Markov's Inequality
yes thats what i meant. thank you very much :D