# Math Help - More p.m.f [p.d.f.] questions!

1. ## More p.m.f [p.d.f.] questions!

I think I have the hang of this now, but I have a minor question:

I am to find the constant $c$ which would make $f(x)$ satisfy all the conditions of being a p.m.f.

$f(x)=c(1/4)^x;~~~x=1,2,3,...$

I assumed that this would continue on to infinity [for there was another similar to this, where it explicitly said that $x=1,2,3...n$]

So, I figured that $f(x)>0$;

I also noticed that $\sum_{x\in S}f(x)=1\implies \sum_{x=1}^{\infty}c(1/4)^x=1$. The series is geometric, where $a=c$, and $r=\frac{1}{4}$. Thus, it converges to $\frac{c}{1-\frac{1}{4}}=\frac{c}{\frac{3}{4}}=\frac{4}{3}c$

So for $\frac{4}{3}c=1,~c=\frac{3}{4}$

However, my text tells me that the answer is 3.

Maybe I'm overlooking one simple thing...wait!

How do I apply the last property of a p.m.f:

$P(X\in A)=\sum_{x\in A}f(x)~~~\text{where}~A\subset S$

to this problem?

I'd appreciate any input!

--Chris

2. $\sum\limits_{k = 1}^\infty {\left( {1/4} \right)^k } = \frac{{(1/4)}}
{{1 - (1/4)}} = \frac{1} {3}\quad \Rightarrow \quad c = 3$

3. Yo Chris...

Remember it's ${\color{red}1}+r+r^2+\dots+r^n+\dots$ that equals $\frac{1}{1-r}$

If it starts at $r$ (which is the case here since the sum starts at 1), we have : $r+r^2+\dots+r^n+\dots=r \left(1+r+r^2+\dots+r^n+\dots\right)={\color{red}r } \cdot \frac{1}{1-r}$

Go back to high school !

4. Originally Posted by Moo
Yo Chris...

Remember it's ${\color{red}1}+r+r^2+\dots+r^n+\dots$ that equals $\frac{1}{1-r}$

If it starts at $r$ (which is the case here since the sum starts at 1), we have : $r+r^2+\dots+r^n+\dots=r \left(1+r+r^2+\dots+r^n+\dots\right)={\color{red}r } \cdot \frac{1}{1-r}$
Ahaha...I can't believe I didn't see that...

Go back to high school !
Never!!

--Chris