Finding a pdf from a CDF with a Discrete Random Variable

Question

I know this question isn't very difficult but I'm not convinced I'm doing it right.

For a discrete random variable if you have the CDF, the pdf is defined as $f(x)=F(x)-F(x-)$.

I have:

$$F(x) =\begin{cases}0,&{x\le 0}\\log(x+1),&{x=1,2,3,...,9}\\1,&{x\gt 9}\\\end{cases}$$

What I did was:

$$f(x)=F(x)-F(x-)$$ $$=log(x+1)-log(x)$$ $$=log\left(\frac{x+1}{x}\right)$$

Making the pdf: $f(x) =log\left(\frac{x+1}{x}\right)$ for $x=1,2,3,...,9$

This was so trivial I'm in the mind set that I must be missing some detail. Is this correct?

It doesn't say. I don't know if it is correct to do so, but I was once told to assume base 10 if it wasn't indicated. — Vincent, Jul 16 '14 at 10:05

score 1 · Accepted Answer · answered Jul 16 '14 at 10:08

1

Your solution is correct. You can check this by the telescoping sum: $\sum_{k=1}^{9} \log_{10} (\frac{k+1}{k}) = \log_{10}10 = 1$. Hence this is a probability distribution and each $0<p_k<1$

answered Jul 16 '14 at 10:08

Alex

19,262

Thanks for checking it out. – Vincent Jul 16 '14 at 10:19
you are welcome. Logarithm must be base 10 though. – Alex Jul 16 '14 at 10:20

Finding a pdf from a CDF with a Discrete Random Variable

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