MHB Geometric rv and exponential rv question

AI Thread Summary
The discussion focuses on calculating the probability for a geometric random variable (rv) and an exponential rv. The user initially attempts to find the probability using the probability density function (PDF) but realizes they may be on the wrong track. The correct approach involves using the cumulative distribution function (CDF) to find the probability \( P(k-1 \leq X < k) \). This is achieved by subtracting the CDF values at the specified points, leading to a simplified expression after canceling terms. The conversation highlights the importance of using the CDF for this type of calculation.
nacho-man
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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!
 

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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 $e^{-\lambda(k-1)}$.
 
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