Can someone please explain this to me?

[mdavis567]

This problem is giving me a solution that I simply don't understand...

The pf is P[X=x] = 2/3^x, and I'm supposed to find the probability that X is even.

I get that its the sum of the probabilities of X = 2,4,6...but what I don't get is how the book is getting this:

P[X=2] + P[X=4] +... = 2/3 * (1/3 + 1/3^3 + 1/3^5+....) = 2/3^2 * 1/(1-(1/3^2)) = 1/4

I'm not following the simplification, how do they get the 2nd part from the first, and the 3rd from the 2nd?

[jprep]

This is a geometric series.

P[X=2] = 2/3^2
P[X=4] = 2/3^4
P[X=6] = 2/3^6 and so on.

Now you add all these up. That's

2/3^2 + 2/3^4 + 2/3^6 + 2/3^8 + ...

Now factor the first term out of every term (factor out 2/3^2, that is), so that you get this:

2/3^2 * (1 + 1/3^2 + 1/3^4 + 1/3^6 + ...) which is equivalent to:

2/9 * (1 + [1/9]^1 + [1/9]^2 + [1/9]^3 + ...)

The bolded part is special, here's the rule (yes, you'll just have to memorize this, there is a proof though):

(1 + [1/r]^1 + [1/r]^2 + [1/r]^3 + ...) = 1 / (1 - r)

In your question, r is 1/9. So the above bolded part becomes:

2/9 * (1 / (1 - 1/9)) = 2/9 * (1 / (8/9)) = 2/9 * 9/8 = 2/8 = 1/4


It's not supposed to actually look that hard, I just don't know how to use the Tex thingy on this forum. These kind of geometric series show up a lot in the practice exams, but they don't ask many questions like this on the real exam - they could, though. That problem was solved weirdly - they skipped some algebra, and the second step looks weird.

Good luck!

[mdavis567]

You cleared that right up for me.

As for the clarity, im kind of a dunce as well, and it was fine. I copied it down so it looked nicer, thanks!

[Actuarialsuck]

jprep: http://www.actuarialoutpost.com/actu...ad.php?t=74727

A few key commands:

\frac{1}{2} will produce

\int_{-\infty}^{\infty} will produce

\sum_{i=0}^{n} will produce

You can usually hit quote if you see anyone writing in TeX and see what "code" they used.

[jprep]

Originally Posted by Actuarialsuck

jprep: http://www.actuarialoutpost.com/actu...ad.php?t=74727

A few key commands:

\frac{1}{2} will produce

\int_{-\infty}^{\infty} will produce

\sum_{i=0}^{n} will produce

You can usually hit quote if you see anyone writing in TeX and see what "code" they used.

Thanks! Now that's

Error: Calculation too lame.

[atomic]

Originally Posted by mdavis567

This problem is giving me a solution that I simply don't understand...

The pf is P[X=x] = 2/3^x, and I'm supposed to find the probability that X is even.

I get that its the sum of the probabilities of X = 2,4,6...but what I don't get is how the book is getting this:

P[X=2] + P[X=4] +... = 2/3 * (1/3 + 1/3^3 + 1/3^5+....) = 2/3^2 * 1/(1-(1/3^2)) = 1/4

I'm not following the simplification, how do they get the 2nd part from the first, and the 3rd from the 2nd?

A simpler approach would be to observe that since



in order for the given mass function to correspond to a valid probability function, we see that

.

Therefore the probability that X is even is simply 1/3 the probability that X is odd, and because the probability that X is either even or odd is 1, we obtain

.

[Colymbosathon ecplecticos]

Cute!