Probability density,cumulative function problem

AI Thread Summary
The discussion centers on solving a probability density function (PDF) for a telephone service center, specifically f(x) = c(2x-1)/(5x), where X represents the number of tries (phone calls). Participants are attempting to determine the constant c and the cumulative distribution function (CDF). Initial calculations led to conflicting values for c, with one user suggesting c = 3/4 and another correcting it to c = 6/5 based on the geometric series sum. The importance of summing from 0 to infinity, given that negative phone calls are not feasible, is emphasized. The final simplified PDF is presented as f(x) = (3/5)(2/5)^x.
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Homework Statement


Prob. density function is given below, representing probability of answer in a telephone service center
f(x) = c 2x-1/ 5x

where X is the random variable representing tries (phone calls).
a)find c
b)find the cumulative distribution function
i think i can solve the rest if i get through the first obstacles.

Homework Equations


The sum of a discrete density function from -inf to inf equals 1



The Attempt at a Solution


X is discrete, and the sum from -inf to inf must equal 1.
I sum up from 0 to inf (as negative phone calls don't make sense. is that right to do?) and come up with c=3/4
SUM c 2x+1 /5x , from 0 to inf
=lim, k->inf of SUM (1/3)x , x from 0 to k
=((c/2) * 1-(1/3)inf + 1 )/(2/3) = 1
=>c=4/3

So f is now simplified to 1/3x+1

Is the above method correct? Sadly, i then take the sum from 0 to inf to verify that it equals 1, but it doesn't.
How can i find the cumulative distribution?
 
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atrus_ovis said:

The Attempt at a Solution


X is discrete, and the sum from -inf to inf must equal 1.
I sum up from 0 to inf (as negative phone calls don't make sense. is that right to do?) and come up with c=3/4
SUM c 2x+1 /5x , from 0 to inf
=lim, k->inf of SUM (1/3)x , x from 0 to k
=((c/2) * 1-(1/3)inf + 1 )/(2/3) = 1
=>c=4/3

So f is now simplified to 1/3x+1

You seem to have multiple mistakes here. The sum is

c \sum_{x=0}^\infty \frac{2^{x-1}}{5^x} = \frac{c}{2} \sum_{x=0}^\infty \left(\frac{2}{5}\right)^x

Also 2/5\neq 1/3. You want to apply the formula for a geometric sum directly to the expression with 2/5 in it.
 
Ah, ok.
Then c = 6/5, and f(x) is simplified to (3/5 )* (2/5)x
 
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