Let the random variable X have pmf ... and let Y=1/X. Find the cdf of Y.

Question

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!

Answer 1

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 $\lceil z\rceil$.

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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 $2^{1-\lceil\frac1x\rceil}$