Prove that if X∼ Geometric (p) then, E(X)=q/p, Var(X)=q/p^2, m_X(t)=p(1-qe^t).

Jensen Mclean 2022-09-03 Answered
Prove that if X Geometric (p) then, E ( X ) = q p Var ( X ) = q p 2 m X ( t ) = p ( 1 q e t )
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Answers (2)

Belen Solomon
Answered 2022-09-04 Author has 5 answers
Step 1
For the variance, it suffices to compute E [ X 2 ]. You can adapt the approach you used for computing E[X]; it may help to manipulate the series for 1 ( 1 x ) 3 .
Here is a direct derivation:
E [ X 2 ] = k 0 k 2 p q k ( 1 q ) E [ X 2 ] = E [ X 2 ] q E [ X 2 ] = k 0 k 2 p q k m 0 m 2 p q m + 1 = k 0 k 2 p q k k 1 ( k 1 ) 2 p q k = k 1 ( 2 k 1 ) p q k = 2 k 1 k p q k k 1 p q k = 2 E [ X ] ( 1 p ) = 2 q p q = q ( 2 p ) p .
Given the above expression for ( 1 q ) E [ X 2 ], you can then do a few more steps compute the variance of X.
Step 2
For the MGF, m X ( t ) = E [ e t X ] = k 0 e t k p q k = p k 0 ( e t q ) k which is a geometric series that you should already know how to compute. By the way, there is a typo in your expression for the MGF.
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Austin Rangel
Answered 2022-09-05 Author has 2 answers
Step 1
I would do the MGF calculation first, then use the result to compute the expectation and variance. That's probably not the intent of the problem, but it is mathematically valid.
M X ( t ) = E [ e t X ] = x = 0 e t x ( 1 p ) x p .
Step 2
Then, E [ X ] = M X ( 0 ) , E [ X 2 ] = M X ( 0 ) ,, and Var [ X ] = E [ X 2 ] E [ X ] 2 .
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