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I'm trying to solve a problem on Prasanna Sahoo's book on Probability and Statistics. One problem assumes that $X$ is a random variable with probability density function $$f(x)=\frac{2c}{3^{x}}$$ for $x=1,2,3,...,\infty$ where $c$ is constant. According to the answer key, the value of $c$ is 1. Since $x$ is countable, then an infinite series must be used. Also, by a theorem, its summation must be equal to 1. So we have the equation $$\sum_{x=1}^{\infty}\frac{2c}{3^{x}}=1.$$ The summation above is a geometric series. Hence we can use the formula $$\sum_{n=1}^{\infty}ar^{n}=\frac{a}{1-r}.$$ However, $$\sum_{x=1}^{\infty}\frac{2c}{3^{x}}=\sum_{x=1}^{\infty}2c\Big(\frac{1}{3}\Big)^{x}=\frac{2c}{1-1/3}=1.$$ Thus, $c=1/3$. This does not coincide with the answer stated at the back of the book. One solution that I thought of is by changing the index of the summation, as follows: $$\sum_{x=1}^{\infty}\frac{2c}{3^{x}}=\sum_{x=0}^{\infty}\frac{2}{3}\Big(\frac{1}{3}\Big)^x.$$ From here, we can let $a=2/3$ and $r=1/3$, which will make $c=1$. However, I can't justify the change in the index of summation. Some textbooks use $n=1$ in the summation of an infinite geometric series, while some use $n=0$. I haven't found an article stating the difference between the two indexes when considering a probability density function.

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You missed the first term in the formula of sum of decreasing GP is: $$S=\color{red}{\frac13}+\frac{1}{3^2}+\cdots =\frac{\color{red}{\frac13}}{1-\frac13}=\frac12.$$ Hence $c=1$.

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Hint:

$$\sum_{x=1}^n r^x=r\cdot \frac{1-r^n}{1-r}=r\cdot \left(\frac{1}{1-r}-\frac{r^n}{1-r} \right)$$

For $|r|<1$

$$\lim_{n\to \infty}r\cdot \left(\frac{1}{1-r}-\frac{r^n}{1-r} \right)= \frac{r}{1-r}$$


Another (classical) version of the partial sum is

$$\sum_{x=\color{red}0}^{\color{red}{n-1}} r^x= \frac{1-r^n}{1-r}=\frac{1}{1-r}-\frac{r^n}{1-r} $$

For $|r|<1$

$$\lim_{n\to \infty} \frac{1}{1-r}-\frac{r^n}{1-r} = \frac{1}{1-r}$$

The different of the two versions are the upper bounds and lower bounds of the partial sums. In both cases the number of summands are $n$

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  • $\begingroup$ Why did you use the formula for partial sums instead of an infinite series? There's an obvious difference, but I honestly don't understand the difference. $\endgroup$ – Liam Ceasar Feb 22 '18 at 10:20
  • $\begingroup$ @LiamCeasar I have used the partial sum formula to show how the formula of the infinite case is derived. Yes there is a difference what you have posted, which was not right. Please ask a specific question what do you not understand about my answer. I will give an reply. $\endgroup$ – callculus Feb 22 '18 at 10:32
  • $\begingroup$ I mean why was r/(1-r) derived instead of using the classic formula a/(1-r) for an infinite series? $\endgroup$ – Liam Ceasar Feb 22 '18 at 10:38
  • $\begingroup$ @LiamCeasar I will make an edit. $\endgroup$ – callculus Feb 22 '18 at 10:40
  • $\begingroup$ @LiamCeasar I´ve finished the edit. $\endgroup$ – callculus Feb 22 '18 at 10:48

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