# Talk:Moment Generating Function of Geometric Distribution/Formulation 1/Examples/First Moment

Jump to navigation
Jump to search

I get this:

\(\ds \map { {M_X}'} t\) | \(=\) | \(\ds \map {\frac \d {\d t} } {\dfrac {1 - p} {1 - p e^t} }\) | Moment Generating Function of Geometric Distribution | |||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac {-\paren {1 - p} \paren {-p e^t } } {\paren {1 - p e^t}^2 }\) | Quotient Rule for Derivatives, Derivative of Exponential Function | |||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac {\paren {1 - p} \paren {p e^t} } {\paren {1 - p e^t}^2 }\) | get rid of all that confusing negativeness | |||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac {p e^t - p^2 e^t} {\paren {1 - p e^t}^2}\) |

which is not the same as $\dfrac {p e^t } {\paren {1 - p e^t} }$

What am I doing wrong here? I can't see my mistake. --prime mover (talk) 14:45, 20 April 2021 (UTC)

- My error - will correct. --Robkahn131 (talk) 15:23, 20 April 2021 (UTC)
- No wait - it's ok - $\paren {1 - p e^t}$ cancels
- Nope - it's wrong - will fix. --Robkahn131 (talk) 15:26, 20 April 2021 (UTC)