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