00:01
In the question it is given that here we have to show that by use the mgf for the geometric or the geometric random variable, random variable, variable.
00:53
So e bracket x is equal to p and v a r bracket x is equal to q upon p square, q upon p square.
01:10
So we can say that here we know that the moment generating function mgf, mgf is m bracket t, m bracket t is equal to p e t square upon 1 minus q e square t where p, where p is equal to probability of success, probability, probability of success.
02:28
P is equal to probability of success and q is equal to probability of failure, q is equal to probability of failure, probability of failure, of failure.
02:48
So we know that, we know that q plus, p plus q is equal to 1, p plus q is equal to 1 and e and e bracket x is equal to d upon dx, sorry d upon dt, we can say that m bracket t, m bracket t, t is equal to 0.
03:30
Now we know that v a r bracket x is equal to m bracket x is equal to m double differentiation, m double differentiation bracket 0 minus m single differentiation bracket 0 square.
03:57
Now we can say that thus, thus, thus m single differentiation bracket t is equal to d upon dt bracket p e t upon 1 minus q e and t square bracket close.
04:36
Now we know that 1 minus q e, it will be 1 minus q e t bracket and bracket p e t minus p e square t bracket minus q e square t, 1 upon 1 minus q e t square.
05:12
Now when we solved it, what we get? we get that it will be equal to p e square t minus p q e, p q e square 2 t plus p q e square 2 t upon 1 minus q e t whole square.
05:50
And now what we get? we get that will be equal to p e square t upon 1 minus q e square t whole square.
06:08
Now it will be, we know that m single differentiation bracket t where t is equal to 0 will be equal to p e square t upon 1 minus q e square t whole square where t is equal to 0.
06:40
So, p upon 1 minus q whole square where p upon bracket x minus x plus p whole square and that will be equal to, that will be equal to p, that will be equal to p upon p square since we know that since p plus q is equal to 1, so it will be 1 upon p and that will be equal to, so hence it is proved that e bracket x that will be equal to 1 upon p.
07:38
So, first it will be our answer.
07:42
First it will be our answer.
07:45
Now what we can do? now we know that, now we know that m double differentiation, m double differentiation bracket t is equal to d upon dt bracket p e square t upon 1 minus q e square t whole square...