MHB Why is my solution for part b) of the geometric series question incorrect?

[nacho-man]

Please refer to the attached sheet.
I need help with part b)

for part
a)
I did:
$\sum\limits_{n=0}^{\infty} a^n = \frac{1}{1-a}$
So for

$\sum\limits_{x=0}^{\infty} a^{2x}$
$a^{2x} = (a^2)^x$
and

$\sum\limits_{x=0}^{\infty} (a^2)^x = \frac{1}{1-a^2}$
for part b)
the solutions say i am wrong.

I did this:

$c$ $\sum\limits_{n=0}^{\infty} a^n = \frac{c}{1-a} = 1$
therefore
$c = 1-a^2$ (from part a)

The answer is not this. Solutions are:$\sum\limits_{x=1}^{\infty} c(a^2)^x = 1$
$ ...$
$c = \frac{a^{2x}(1-a^2)}{a^2}$

Why do we begin from $x=1$ and not $x=0$ when summing this series? Is this always the case for these questions? How do I solve part b?

 

[chisigma]

Please refer to the attached sheet.
I need help with part b)

for part
a)
I did:
$\sum\limits_{n=0}^{\infty} a^n = \frac{1}{1-a}$
So for

$\sum\limits_{x=0}^{\infty} a^{2x}$
$a^{2x} = (a^2)^x$
and

$\sum\limits_{x=0}^{\infty} (a^2)^x = \frac{1}{1-a^2}$
for part b)
the solutions say i am wrong.

I did this:

$c$ $\sum\limits_{n=0}^{\infty} a^n = \frac{c}{1-a} = 1$
therefore
$c = 1-a^2$ (from part a)

The answer is not this. Solutions are:$\sum\limits_{x=1}^{\infty} c(a^2)^x = 1$
$ ...$
$c = \frac{a^{2x}(1-a^2)}{a^2}$

Why do we begin from $x=1$ and not $x=0$ when summing this series? Is this always the case for these questions? How do I solve part b?

Using the n instead of the x we have...

$\displaystyle p_{X}(n) = c\ a^{2\ n},\ 0 < a < 1,\ n = 1,2,...\ (1)$

The value of c is computed from the basic relation...

$\displaystyle \sum_{n=1}^{\infty} p_{X} (n) = 1 \implies c = \frac{1 - a^{2}}{a^{2}}\ (2)$

The mean value is...

$\displaystyle \mu_{X} = c\ \sum_{n=1}^{\infty} n\ a^{2 n} = \frac{1}{1-a^{2}}\ (3)$

Are You able to proceed computing variance?...

Kind regards

$\chi$ $\sigma$

 

[nacho-man]

I can find variance.

This is what I don't understand though.

The value of c is computed from the basic relation...

$\displaystyle \sum_{n=1}^{\infty} p_{X} (n) = 1 \implies c = \frac{1 - a^{2}}{a^{2}}\ $

Why do we some from $n=1$ and not $n=0$ ? If I had a different series, would I take the same approach?

 

[chisigma]

I can find variance.

This is what I don't understand though.

The value of c is computed from the basic relation...

$\displaystyle \sum_{n=1}^{\infty} p_{X} (n) = 1 \implies c = \frac{1 - a^{2}}{a^{2}}\ $

Why do we some from $n=1$ and not $n=0$ ? If I had a different series, would I take the same approach?

The definition seems to indicate that $p_{X} (0) = 0$ so that the series starts from n=1. The variance can be found first computing... $\displaystyle E \{ X^{2}\} = \sum_{n=1}^{\infty} n^{2}\ p_{X} (n) = c \sum_{n=1}^{\infty} n^{2} a^{2 n} = \frac{1 + a^{2}}{(1 - a^{2})^{2}}\ (1)$

... so that is...

$\displaystyle \sigma_{X}^{2} = E \{ X^{2}\} - E^{2} \{ X \} = \frac{a^{2}}{(1 - a^{2})^{2}}\ (2)$ Kind regards $\chi$ $\sigma$

 

[nacho-man]

The definition seems to indicate that $p_{X} (0) = 0$ so that the series starts from n=1.

Although,
if $p_X(x)$ = $c a^{2x}$
Doesn't $p_X(0) = c$ ?
and $ c \neq 0$
I still don't understand.

 

[chisigma]

Although,
if $p_X(x)$ = $c a^{2x}$
Doesn't $p_X(0) = c$ ?
and $ c \neq 0$
I still don't understand.

The PDF of the discrete variable X is defined as...

$\displaystyle p_{X} (n) = c\ a^{2 n},\ 0 < a < 1,\ n=1,2,... \ (1)$

... and that means that X may be equal to 1,2,... but not equal to 0...

Kind regards

$\chi$ $\sigma$

 

[nacho-man]

The PDF of the discrete variable X is defined as...

$\displaystyle p_{X} (n) = c\ a^{2 n},\ 0 < a < 1,\ n=1,2,... \ (1)$

... and that means that X may be equal to 1,2,... but not equal to 0...

Kind regards

$\chi$ $\sigma$

wow thank you, i can't believe i missed that...

(Doh)