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Math Help - Statistics - Geometric Distribution

  1. #1
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    Statistics - Geometric Distribution

    Suppose X has a geometric distribution with success probability, 1/3.

    Determine the probability distribution of Y = X/(X+1) that provides the possible values of Y and its probability mass function.


    Basically, I am terrible with statistics and any help that could lead me into the right direction for solving this problem will be greatly appreciated.
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    Re: Statistics - Geometric Distribution

    solving this is the same as the other one.

    $f_X(x)=(1-p)^x p,~~0\leq x$

    find $F_Y(y)=Pr[Y<y]=Pr[\frac{X}{X+1}<y]$

    then $f_Y(y)=\frac{\partial}{\partial y}F_Y(y)$
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  3. #3
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    Re: Statistics - Geometric Distribution

    For this, I did:

    F(y) = P[Y < y] = P[ x/(x+1) < y] = P [ x < y(x+1) ]

    I've been trying to get the x's on one side so that I could do P[ x < whatever this would be after I solved for x] = the integral from 0 to whatever this is

    That's not a really good explanation, but I think I'm on the right track with this problem?
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    Re: Statistics - Geometric Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    For this, I did:

    F(y) = P[Y < y] = P[ x/(x+1) < y] = P [ x < y(x+1) ]

    I've been trying to get the x's on one side so that I could do P[ x < whatever this would be after I solved for x] = the integral from 0 to whatever this is

    That's not a really good explanation, but I think I'm on the right track with this problem?
    come on...

    first off note that $X \geq 0$ so $X+1 > 0$

    $\dfrac{X}{X+1}<Y$

    $X<(X+1)Y$

    $X(1-Y)<Y$

    now we have to be careful. Let's find out what values Y can take.

    $X=k, k \in \mathbb{I}, k\geq 0$ so

    $Y=\dfrac{X}{X+1} = \{0, \frac{1}{2}, \frac{2}{3}, \dots\}$ and we note $0 \leq Y < 1$

    so $0<1-Y\leq 1$ and we can divide by $1-Y$ without flipping the inequality sign. So

    $X<\dfrac{Y}{1-Y}$

    so $Pr[Y<y]=Pr[X<\dfrac{y}{1-y}], 0\leq y < 1$

    see if you can finish it from here.
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  5. #5
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    Re: Statistics - Geometric Distribution

    This is what I got:

    F(y) = P[Y<y] = P[ x/(x+1) < y] = P[ x < y/(1-y) ] = integral from 0 to y/(1-y) (1 - p)^x * p dx = [p*((1-p)^(y/1-y) - 1] / ln(1-p)
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    Re: Statistics - Geometric Distribution

    Was that integral above correct? Or would the integral actually be:

    F(y) = P[ x < y/(1-y) ] = integral from 0 to 1 (1-p)^x * p dy

    But, that doesn't seem correct either.
    It's so frustrating not getting this.
    Any help/feedback is really appreciated
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  7. #7
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    Re: Statistics - Geometric Distribution

    I think you're making this harder than it is.

    You know, or can look up the CDF of the geometric distribution. Just get that and plug $\dfrac{y}{1-y}$ in for $k$.
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    Re: Statistics - Geometric Distribution

    I'd rather not do the c.d.f. if the p.d.f. would be easier. You said that I'm making this harder than it really is, so I tried to not over think as much and I kept getting this:

    F(y) = the integral from 0 to y/1-y of f(x) dx

    I know that the derivative of F(y) will give me f(y), but it's the set up of the integral that's getting me. As soon as I can figure that out, I know I can easily do the rest.
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    Re: Statistics - Geometric Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    I'd rather not do the c.d.f. if the p.d.f. would be easier. You said that I'm making this harder than it really is, so I tried to not over think as much and I kept getting this:

    F(y) = the integral from 0 to y/1-y of f(x) dx

    I know that the derivative of F(y) will give me f(y), but it's the set up of the integral that's getting me. As soon as I can figure that out, I know I can easily do the rest.
    wat?

    $F_X(k)=1 - (1-p)^{k+1}$

    $F_Y(y)=1 - (1-p)^\left({\frac{y}{1-y}+1}\right)$

    I forget what we're doing here but if you need $f_Y(y)$ just take the derivative w/respect to $y$.
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    Re: Statistics - Geometric Distribution

    For my answer, I got:

    f(y) = (1 - p)^(1 - y) * ln(1 - p) for 0 <= y < 1 and 0 elsewhere
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    Re: Statistics - Geometric Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    For my answer, I got:

    f(y) = (1 - p)^(1 - y) * ln(1 - p) for 0 <= y < 1 and 0 elsewhere
    I work out that

    $\large F_Y(y,p)=1-(1-p)^{\frac{y}{1-y}+1}$

    and

    $\large f_Y(y,p)=\frac{(p-1) (1-p)^{\frac{y}{1-y}} \log (1-p)}{(y-1)^2}$

    for $0 \leq y < 1$

    Statistics - Geometric Distribution-clipboard01.jpg
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  12. #12
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    Re: Statistics - Geometric Distribution

    Thinking about this a bit further I think you have to use the the floor function on $\dfrac{y}{1-y}$ since that exponent really needs to be an integer to make sense as the CDF of an exponential distribution. So you'd have

    $F_Y(y,p)=1-(1-p)^{\left \lfloor \frac{y}{1-y}\right \rfloor+1}$

    $f_Y(y,p)$ is going to be a staircase sort of function
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  13. #13
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    Re: Statistics - Geometric Distribution

    I screwed this whole thing up about as much as possible.

    Really all you need to do here is calculate for each $X$, what $Y=\frac{X}{X+1}$ is and assign X's probability to that value.

    for example for $X=0, Y=1$ and has probabilty $p$. Similarly in the chart below for $p=\frac{1}{3}$

    $
    \begin{array}{cccc}
    X &Y &p_Y(y) &F_Y(y) \\
    0 & 0 & 0.333333 & 0.333333 \\
    1 & \frac{1}{2} & 0.222222 & 0.555556 \\
    2 & \frac{2}{3} & 0.148148 & 0.703704 \\
    3 & \frac{3}{4} & 0.0987654 & 0.802469 \\
    4 & \frac{4}{5} & 0.0658436 & 0.868313 \\
    5 & \frac{5}{6} & 0.0438957 & 0.912209 \\
    6 & \frac{6}{7} & 0.0292638 & 0.941472 \\
    7 & \frac{7}{8} & 0.0195092 & 0.960982 \\
    8 & \frac{8}{9} & 0.0130061 & 0.973988 \\
    9 & \frac{9}{10} & 0.00867076 & 0.982658 \\
    10 & \frac{10}{11} & 0.00578051 & 0.988439 \\
    11 & \frac{11}{12} & 0.00385367 & 0.992293 \\
    12 & \frac{12}{13} & 0.00256912 & 0.994862 \\
    13 & \frac{13}{14} & 0.00171274 & 0.996575 \\
    14 & \frac{14}{15} & 0.00114183 & 0.997716 \\
    15 & \frac{15}{16} & 0.000761219 & 0.998478 \\
    \end{array}$
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  14. #14
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    Re: Statistics - Geometric Distribution

    How did you know to use those values for both X and Y in the chart?
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    Re: Statistics - Geometric Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    How did you know to use those values for both X and Y in the chart?
    X is geometrically distributed so X=0,1,2 .... I just stopped at 15 because there's not room to list the whole infinite sequence.

    $Y=\frac{X}{X+1}$, these come from letting X be as above.
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