by a discrete!?

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July 9th 2008, 08:42 AM
alv

how a continous can be added/subtracted with/by a discrete!?

hello world! this is my first post to this forum.

please answer to the following seemingly classical probability question, it's urgent for me:

suppose X is a Poisson(1) random variable, and Z is a Standard Normal {N(0,1)} random variable. what is the result of the following probability:

P(X > (16 - Z^4) / (X + Z^2)) < 3/8

PS. by the sign ^ i mean power, so e.g. Z^4 is Z to the power of four.
July 9th 2008, 08:52 AM
meymathis

Hi,

What do you mean by "what is the result of the following probability"? Are you asking whether the statement is true or false? Or are you being asked to prove it?
July 9th 2008, 11:00 PM
alv

yes meymathis is right...

Quote:

Originally Posted by meymathis

Hi,

What do you mean by "what is the result of the following probability"? Are you asking whether the statement is true or false? Or are you being asked to prove it?

oops! yes you're right meymathis! i meant to prove that inequality. so sorry for this mistake...waiting for your replies
July 10th 2008, 12:51 PM
meymathis

Sorry this took so long. It took me a while to get my thinker working. Hope this isn't too late!? Also, I proved something ever so slightly weaker than what you have here. I hope that there is a typo and it should have been $\leq3/8$ (Worried)

Whenever I see a proof using with an inequality, the first thoughts through my head is Markov, Chebyshev, or Jensen's inequalities. Let's start with the most obvious, Markov. Are you allowed to use this one?

$ \mathbf{P}\left( X>\frac{16-Z^4}{X+Z^2}\right) $

$ =\mathbf{P}\left( X^2+XZ^2+Z^4>16\right) $

Now use Markov.

$ \leq \frac{\mathbf{E}\left[ X^2+XZ^2+Z^4\right] }{16} $

$ =\frac{\mathbf{E}\left[ X^2\right]+\mathbf{E}\left[XZ^2\right]+\mathbf{E}\left[Z^4\right] }{16} $

Now by independence:

$ =\frac{\mathbf{E}\left[ X^2\right]+\mathbf{E}\left[X\right]\mathbf{E}\left[Z^2\right]+\mathbf{E}\left[Z^4\right]}{16} $

X is Poisson(1) (Poisson Distribution -- from Wolfram MathWorld equationa 19 and 22)so
$\mathbf{E}\left[X\right] = 1$

$\mathbf{E}\left[X^2\right] = 2$

Z is standard normal (http://en.wikipedia.org/wiki/Normal_distribution#Moments) so
$\mathbf{E}\left[Z^2\right] = 1$

$\mathbf{E}\left[Z^4\right] = 3$

That get's it except that its $\leq$ rather than <. (Doh)
July 10th 2008, 12:59 PM
frenzy

Quote:

Originally Posted by meymathis

...
$ =\mathbf{P}\left( X^2+XZ^2+Z^4>16\right) $

Now use Markov.

$ \leq \frac{\mathbf{E}\left[ X^2+XZ^2+Z^4\right] }{16} $

...

That get's it except that its $\leq$ rather than <. (Doh)

Since you have $\mathbf{P}\left( X^2+XZ^2+Z^4>16\right)$ (not $\mathbf{P}\left( X^2+XZ^2+Z^4\geq16\right))$

you get ...

$ =\mathbf{P}\left( X^2+XZ^2+Z^4>16\right)< \frac{\mathbf{E}\left[ X^2+XZ^2+Z^4\right] }{16} $
July 10th 2008, 01:05 PM
meymathis

Thanks Frenzy. Details, details, details....

(Headbang)
July 10th 2008, 01:28 PM
meymathis

Quote:

Originally Posted by frenzy

Since you have $\mathbf{P}\left( X^2+XZ^2+Z^4>16\right)$ (not $\mathbf{P}\left( X^2+XZ^2+Z^4\geq16\right))$

you get ...

$ =\mathbf{P}\left( X^2+XZ^2+Z^4>16\right)< \frac{\mathbf{E}\left[ X^2+XZ^2+Z^4\right] }{16} $

You know actually, I don't believe I've seen this before. Not that its hard, but it does not follow from Markov. The proof is almost exactly the same. Depending on the teacher, they may demand to see the proof of this variation of Markov. See the proof at http://en.wikipedia.org/wiki/Markov's_inequality#Special_case:_probability_theo ry but just adapt the $X\geq a$ to $X>a$