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

  1. #1
    Rhymes with Orange Chris L T521's Avatar
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    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
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  2. #2
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    \sum\limits_{k = 1}^\infty  {\left( {1/4} \right)^k }  = \frac{{(1/4)}}<br />
{{1 - (1/4)}} = \frac{1} {3}\quad  \Rightarrow \quad c = 3
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  3. #3
    Moo
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    A Cute Angle Moo's Avatar
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    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 !
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Moo View Post
    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
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