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.

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 $f_Y(y)=\frac{\partial}{\partial y}F_Y(y)$

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?

Re: Statistics - Geometric Distribution

Quote:

Originally Posted by

**AwesomeHedgehog** 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.

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)

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

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 $k$.

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.

Re: Statistics - Geometric Distribution

Quote:

Originally Posted by

**AwesomeHedgehog** 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 $y$.

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

1 Attachment(s)

Re: Statistics - Geometric Distribution

Quote:

Originally Posted by

**AwesomeHedgehog** 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 $0 \leq y < 1$

Attachment 30294

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

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}$

Re: Statistics - Geometric Distribution

How did you know to use those values for both X and Y in the chart?

Re: Statistics - Geometric Distribution

Quote:

Originally Posted by

**AwesomeHedgehog** 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. :D

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