Question 1198958
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*[tex \Large P(D = j) = \log_{10}\left(\frac{j+1}{j}\right), \ \ \text{ where } \ j \in \{ 1,2,3,4,5,6,7,8,9\}]


*[tex \Large P(D = 1) = \log_{10}\left(\frac{1+1}{1}\right) = \log_{10}\left(2\right)]


*[tex \Large P(D = 2) = \log_{10}\left(\frac{2+1}{2}\right) = \log_{10}\left(\frac{3}{2}\right)]


*[tex \Large P(D = 3) = \log_{10}\left(\frac{3+1}{3}\right) = \log_{10}\left(\frac{4}{3}\right)]


*[tex \Large P(D = 4) = \log_{10}\left(\frac{4+1}{4}\right) = \log_{10}\left(\frac{5}{4}\right)]


and so on.


We have these base 10 logs for terms j=1 through j=9
log(2)
log(3/2)
log(4/3)
log(5/4)
log(6/5)
log(7/6)
log(8/7)
log(9/8)
log(10/9)


Adding up those logs, and then using the rule
log(A)+log(B) = log(A*B)
will get us this
log( 2*(3/2)*(4/3)*(5/4)*(6/5)*(7/6)*(8/7)*(9/8)*(10/9) )


Each denominator cancels out with the previous numerator. 
This means every denominator goes away, and nearly every numerator also goes away.


We'll be left with the single term "10" inside the log.
Then we say log(10) = 1


Therefore, the sum of the probability values is 1, which satisfies one of the requirements for the PMF.


The other requirement is that each P(j) value is between 0 and 1.
0 ≤ log(x) ≤ 1
10^0 ≤ 10^log(x) ≤ 10^1
1 ≤ x ≤ 10
To have P(j) between 0 and 1, it is equivalent to having the argument of the log between 1 and 10.
This applies to each log mentioned, so we have satisfied the second requirement.


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Answer: This is a valid PMF since the P(j) values add to 1, and each P(j) value is between 0 and 1.
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