0

Random variables $X_1,X_2,X_3$ are independent with distributions: $P\{X_1=k\}=1/2^k, P\{X_2=k\}=2/3^k, P\{X_3=k\}=4/5^k,k\in\mathbb N.$ Using characteristic functions, find distribution for random variable $X=X_1+X_2+X_3$.

I will show the solution. However, one step I don't understand - I will note that.

Characteristic functions of variables $X_1,X_2,X_3$ are $$h_{X_1}(t)=\sum_{k=1}^{+\infty}e^{ikt}/2^k=\frac{e^{it}}{2}\frac{1}{1-e^{it}/2}$$ $$h_{X_2}(t)=\sum_{k=1}^{+\infty}2e^{ikt}/3^k=\frac{2e^{it}}{3}\frac{1}{1-e^{it}/3}$$ $$h_{X_3}(t)=\sum_{k=1}^{+\infty}4e^{ikt}/5^k=\frac{4e^{it}}{5}\frac{1}{1-e^{it}/5}$$ $$h_X(t)=h_{X_1+X_2+X_3}(t)=h_{X_1}(t)h_{X_2}(t)h_{X_3}(t)$$ $$=\frac{4}{15}e^{3it}\cdot\frac{1}{(1-e^{it}/2)(1-e^{it}/3)(1-e^{it}/5)}$$ $$=\frac{4}{15}e^{3it}\left(\frac{5}{1-e^{it}/2}-\frac{5}{1-e^{it}/3}+\frac{1}{1-e^{it}/5}\right)$$

Now, in the following step, I don't understand why the sum goes from zero, when in the definition of a problem, $k\in\mathbb N$ - zero is not specified. $$$$ $$=\frac{4}{15}e^{3it}\sum_{k=0}^{+\infty}\left(5\cdot\frac{1}{2^k}-5\cdot\frac{1}{3^k}+\frac{1}{5^k}\right)$$

From here, $$P\{X=k+3\}=\frac{4}{15}\left(5\cdot\frac{1}{2^k}-5\cdot\frac{1}{3^k}+\frac{1}{5^k}\right); k=0,1,2,...$$

Here, $k$ goes from zero because of the previous step.

My question is: Why $k$ goes from zero in the last step?

user7920621
  • 1

1 Answers1

0

You lost $e^{itk}$ when wrote the last step. The formula of geometric series gives $$\frac{1}{1-e^{it}/2}=\sum_{k=0}^{+\infty}e^{itk}/2^k,$$ $$\frac{1}{1-e^{it}/3}=\sum_{k=0}^{+\infty}e^{itk}/3^k,$$ $$\frac{1}{1-e^{it}/5}=\sum_{k=0}^{+\infty}e^{itk}/5^k,$$ and we can write \begin{align} h_X(t)& = \frac{4}{15}e^{3it}\left(\frac{5}{1-e^{it}/2}-\frac{5}{1-e^{it}/3}+\frac{1}{1-e^{it}/5}\right) \cr & = \frac{4}{15}e^{3it} \sum_{k=0}^{+\infty}e^{itk}\left(5\cdot\frac{1}{2^k}-5\cdot\frac{1}{3^k}+\frac{1}{5^k}\right) \cr & =\sum_{k=0}^{+\infty}e^{it(k+3)}\frac{4}{15}\left(5\cdot\frac{1}{2^k}-5\cdot\frac{1}{3^k}+\frac{1}{5^k}\right) \cr & = \sum_{k=3}^{+\infty}e^{itk}\frac{4}{15}\left(5\cdot\frac{1}{2^{k-3}}-5\cdot\frac{1}{3^{k-3}}+\frac{1}{5^{k-3}}\right) \end{align} You can see that $X$ can take values $k=3,4,\ldots$ w.p. $$\mathbb P(X=k)=\frac{4}{15}\left(5\cdot\frac{1}{2^{k-3}}-5\cdot\frac{1}{3^{k-3}}+\frac{1}{5^{k-3}}\right). $$ The smallest value $k=3$ appears when $X_1=X_2=X_3=1$.

NCh
  • 13,807
  • 4
  • 17
  • 33