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I would appreciate some help on this problem.

"Let the random variable X have pmf $p_X(k)=\frac{1}{2^k}$ for $k=1,2,...$ and let $Y=\frac{1}{X}$. Find the cdf of $Y$."

This is my solution: $F_Y(x) := P(Y\leqq x) = P(\frac{1}{X}\leqq x) = P(X \geqq \frac{1}{x}) = P(X>\frac{1}{x}) -P(X=\frac{1}{x}) = 1 - P(X\leqq\frac{1}{x}) + P(X=\frac{1}{x}) = 1 - F_X(\frac{1}{x}) + P_X(\frac{1}{x})$.

I now note that (since $X$'s only can take on integer values)

$F_X(x) = P(X\leqq x) = P(X \leqq [x])$ if $[\cdot]$ denotes the rounded of (downward) integer part of $\cdot$.

This is then equal to $\sum_{k=1}^{[x]} P_X(k)=\sum_{k=1}^{[x]} (\frac{1}{2})^k$

And this is later (geometric sum) equal to

$1-(\frac{1}{2})^{[x]-1}$.

So far so good. (I think).

Thus,

$F_Y(x) = 1 + P(X=\frac{1}{x}) - P(X \leqq \frac{1}{x})$

Now I am not sure how to think.

I'm thinking that $P(X=\frac{1}{x}) = p_X(\frac{1}{x}) $ if $x \in Q\setminus (0,1]^c$ OR zero elsewhere.

That is, because $\frac{1}{x} \in \{1,2,3,...\} \implies x \in \{\frac{1}{1},\frac{1}{2},\frac{1}{3},...\}$ because $X$ only can take on positive integers.

And also

$P(X \leqq \frac{1}{x}) = P(X \leqq [\frac{1}{x}]) = F_X(\frac{1}{x})$ if $x \in (0,1]\subseteq $ ℝ and zero elsewhere.

The answer to the question in my text book states that

"If $x=1/n$ for some integer n, then $F_Y(x) = (1/2)^{n-1}$, for other $x \in (0,1]$, $F_Y(x)=(1/2)^{[1/x]}$."

I dont understand what they mean. Is my solution correct, and if not what am i doing wrong? How does my text book's answer translate to mine above?

Thanks!

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  • $\begingroup$ Mistake in calculating geometric sum: $\sum_{k=1}^n2^{-k}=1-2^{-n}$ (so $\neq1-2^{1-n}$) $\endgroup$ – drhab Apr 5 '18 at 7:17
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I haven't overlooked your solution (yet) but will give you some guidelines to solve this.

If a random variable $Y$ is discrete and takes values in some countable set $D$ then - if its CDF must be determined - it is handsome to start with finding $P(Y=d)$ for $d\in D$.

After that we can set: $$F_Y(x)=\sum_{d\in D_x}P(Y=d)\tag1$$ where $D_x=D\cap(-\infty,x]$, and our final step is working out the RHS of $(1)$.

In your case $D:=\{\frac1k\mid k=1,2,\dots\}$ and $P(Y=\frac1k)=P(X=k)=2^{-k}$.

To find $D_x$ we must discern two cases:

  • If $x\leq0$ then $D_x=\varnothing$ and consequently $F_Y(x)=0$.
  • If $x>0$ then $\frac1k\in D_x\iff k\geq\frac1x\geq\lceil\frac1x\rceil$ showing that $D_x=\{\frac1k\mid k\geq\lceil\frac1x\rceil\}$.

where $\lceil z\rceil$ denotes the smallest integer that is not smaller than $z$.

So for $x>0$ based on $(1)$ we find:$$F_Y(x)=\sum_{k=\lceil \frac1x\rceil}^{\infty}2^{-k}=2^{-\lceil \frac1x\rceil}\sum_{k=0}^{\infty}2^{-k}=2^{-\lceil \frac1x\rceil}\cdot2=2^{1-\lceil \frac1x\rceil}$$

The answer in your textbook is not based on $\lceil z\rceil$ but is based on $\lfloor z\rfloor$ (and uses the notation $[z]$ for it) which is the largest integer that is not larger than $z$. That's why it must discern between values of $x\in D$ and values $x\notin D$. That is not the case if we work with $\lceil z\rceil$.


edit

There is a small mistake in your answer.

In general $\sum_{k=1}^n2^{-k}=1-2^{-n}$ (so $\neq1-2^{1-n}$) so that (for positive $x$):$$F_X(x)=1-2^{-\lfloor x\rfloor}$$ Note that where you use the notation $[x]$ I use the notation $\lfloor x\rfloor$.

Substituting in $F_Y(x)=1-F_X(\frac1x)+P(X=\frac1x)=1-F_X(\lfloor\frac1x\rfloor)+P(X=\frac1x)$ we find:$$F_Y(x)=1-(1-2^{-\lfloor\frac1x\rfloor})+P\left(X=\frac1x\right)=2^{-\lfloor\frac1x\rfloor}+P\left(X=\frac1x\right)$$ This agrees with the answer in the textbook.

If $\frac1x$ is an integer $n$ then the outcome is $2^{-\lfloor\frac1x\rfloor}+2^{-\lfloor\frac1x\rfloor}=2^{1-\lfloor\frac1x\rfloor}=2^{1-n}$.

If $\frac1x$ is not an integer then the outcome is $2^{-\lfloor\frac1x\rfloor}$.

Also it agrees with my outcome $2^{1-\lceil\frac1x\rceil}$

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  • $\begingroup$ Thanks!! That makes perfect sense. $\endgroup$ – tarkovsky123 Apr 5 '18 at 9:29
  • $\begingroup$ Glad to help. You are welcome. Btw, the "integer part function" is also known as "floor function", and the function used by me is known as "ceiling function". It has profit to be accustomed with both. $\endgroup$ – drhab Apr 5 '18 at 9:51

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