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Geometric rv and exponential rv question


Active member
Sep 10, 2013
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!



Well-known member
MHB Math Scholar
Jan 30, 2012
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 $e^{-\lambda(k-1)}$.