Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$X$ is a discrete random variable taking on the values $X = 1,3,3^2,3^3,\dots,3^m$ and $f(x) = P(X=x)=\dfrac c x$ for a constant $c$. Find $c$.

Solution: Since $P(X)=1$, we know that $\dfrac c x=1$, so $c=x$. To find x, we have $x = \sum_0^m 3^m$. Since this series summation diverges to infinity, $c = \infty$. This is a fascinating problem, however, something doesn't seem right ... in other words, how can $x = \infty$? Is the first statement $c=x$ incorrect?

share|cite|improve this question

I think you're making a number of erroneous assumptions. You have $f(x)=P(X=x)=c/x$ is the probability distribution of $X$ and the state space is $1,3,\ldots,3^m$. It seems like the problem is asking you to find the normalization $c$ for the probability distribution. To wit, you must have that:

$1=\sum_{i=0}^m P(X=3^i)=\sum_{i=0}^m \frac{c}{3^i}=c\frac{1-(1/3)^{m+1}}{1-(1/3)}$.

which you can check by a similar argument to Peter's answer. Now solve for $c$.

The point is that you don't want to say that $P(X=x)=c/x=1$ for every $x$, that doesn't make much sense since you'd be assigning probability 1 to each event. It's the sum that should equal 1. As well, take care in interpreting $P(X=x)$, this is lingo for asking the probability of the random variable $X$ being equal to $x$, which the problem says is proportional to $1/x$. The point is $x$ lives in the range of $X$, so it doesn't make much sense to try and solve for $x$.

share|cite|improve this answer

The probability that $X=1$ is $\dfrac{c}{1}$, the probability that $X=3$ is $\dfrac{c}{3}$, and so on. It follows that $$\frac{c}{1}+\frac{c}{3}+\frac{c}{3^2}+\cdots +\frac{c}{3^m}=1.$$ We want to find the sum of the finite geometric series $1+x+x^2+\cdots+x^m$. where $x=1/3$. Let
$$f(m)=1+x+x^2+\cdots+x^m.$$ Note that $$xf(m)=x+x^2+x^3+\cdots+x^{m+1}.$$ Subtract. We get $$(1-x)f(m)=1-x^{m+1},$$ and therefore $$f(m)=\frac{1-x^{m+1}}{1-x}.$$ Finally, put $x=1/3$ and note that $c=1/f(m)$.

Remark: If instead of stopping at $3^m$, we continue forever, then instead of the finite sum of the answer, we have an infinite sum. But since $(1/3)^{n+1}$ approaches $0$ as $n\to\infty$, the infinite geometric series has sum $1/(1-x)=3/2$, and therefore we get $c=2/3$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.