MHB Geometric rv and exponential rv question

nacho-man
Messages
166
Reaction score
0
Please refer to the attached image.for part
a) this is what i did:

$G = k$, $k-1< X < k$

so I substituted $k-1$ and $k$ into the given exponential rv,

this gave me

$\lambda e^{-\lambda(k-1)}$ and $\lambda e^{-\lambda k}$
$= \lambda e^{-\lambda(k-1)} + \lambda e^{-\lambda k}$
But I feel like I am on the wrong track.

This question is really hard for me to comprehend, could someone water it down for me a bit, or help me out?

Thanks in advance!
 

Attachments

  • mth2222 midsemq.png
    mth2222 midsemq.png
    8.7 KB · Views: 87
Physics news on Phys.org
To find $P(k-1\le X<k)$ where $X$ has exponential distribution, you need to subtract the values of CDF, not PDF:
\[
P(k-1\le X<k)=F(k)-F(k-1) =(1-e^{-\lambda k})-(1-e^{-\lambda (k-1)})
\]
After canceling 1's, factor out $e^{-\lambda(k-1)}$.
 
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Thread 'Detail of Diagonalization Lemma'
The following is more or less taken from page 6 of C. Smorynski's "Self-Reference and Modal Logic". (Springer, 1985) (I couldn't get raised brackets to indicate codification (Gödel numbering), so I use a box. The overline is assigning a name. The detail I would like clarification on is in the second step in the last line, where we have an m-overlined, and we substitute the expression for m. Are we saying that the name of a coded term is the same as the coded term? Thanks in advance.
Back
Top