Showing the sum of binomial independent variables follows a binomial distribution using moment generating...
So I'm trying to solve the following problem:
Show that if $X_i$ follows a binomial distribution with $n_i$ trials, and probability of $p_i=p$ for $i = 1,2,3...n$, and the $X_i$ are independent, then $sum_{i=1}^{n}{X_i}$ follows a binomial distribution using moment generating functions.
Here's what I've tried so far:
$M_{sum_{i=1}^{n}}(t) = prod_{i=0}^{n}M_{X_i}(t) = prod_{i=0}^{n}(pe^t + 1 -p)^{n_i}$
I'm not sure where to go from here, or if I've even done everything correctly until this point. Any help will be greatly appreciated. Thanks!
probability probability-distributions binomial-distribution moment-generating-functions
asked Dec 2 at 0:50
BeepBoop
1135
add a comment |
So I'm trying to solve the following problem:
Show that if $X_i$ follows a binomial distribution with $n_i$ trials, and probability of $p_i=p$ for $i = 1,2,3...n$, and the $X_i$ are independent, then $sum_{i=1}^{n}{X_i}$ follows a binomial distribution using moment generating functions.
Here's what I've tried so far:
$M_{sum_{i=1}^{n}}(t) = prod_{i=0}^{n}M_{X_i}(t) = prod_{i=0}^{n}(pe^t + 1 -p)^{n_i}$
I'm not sure where to go from here, or if I've even done everything correctly until this point. Any help will be greatly appreciated. Thanks!
probability probability-distributions binomial-distribution moment-generating-functions
asked Dec 2 at 0:50
BeepBoop
1135
add a comment |
So I'm trying to solve the following problem:
Show that if $X_i$ follows a binomial distribution with $n_i$ trials, and probability of $p_i=p$ for $i = 1,2,3...n$, and the $X_i$ are independent, then $sum_{i=1}^{n}{X_i}$ follows a binomial distribution using moment generating functions.
Here's what I've tried so far:
$M_{sum_{i=1}^{n}}(t) = prod_{i=0}^{n}M_{X_i}(t) = prod_{i=0}^{n}(pe^t + 1 -p)^{n_i}$
I'm not sure where to go from here, or if I've even done everything correctly until this point. Any help will be greatly appreciated. Thanks!
probability probability-distributions binomial-distribution moment-generating-functions
asked Dec 2 at 0:50
BeepBoop
1135
So I'm trying to solve the following problem:
Show that if $X_i$ follows a binomial distribution with $n_i$ trials, and probability of $p_i=p$ for $i = 1,2,3...n$, and the $X_i$ are independent, then $sum_{i=1}^{n}{X_i}$ follows a binomial distribution using moment generating functions.
Here's what I've tried so far:
$M_{sum_{i=1}^{n}}(t) = prod_{i=0}^{n}M_{X_i}(t) = prod_{i=0}^{n}(pe^t + 1 -p)^{n_i}$
I'm not sure where to go from here, or if I've even done everything correctly until this point. Any help will be greatly appreciated. Thanks!
probability probability-distributions binomial-distribution moment-generating-functions
probability probability-distributions binomial-distribution moment-generating-functions
asked Dec 2 at 0:50
BeepBoop
1135
asked Dec 2 at 0:50
BeepBoop
1135
asked Dec 2 at 0:50
BeepBoop
1135
asked Dec 2 at 0:50
BeepBoop
1135
asked Dec 2 at 0:50
BeepBoop
1135
1135
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
begin{align}
M_{sum_{i=1}^{n}X_i}(t) &= prod_{i=1}^{n}M_{X_i}(t)\
& = prod_{i=1}^{n}(pe^t + 1 -p)^{n_i}\
&=(pe^t + 1 -p)^{sum_{i=1}^{n} n_i}
end{align}
answered Dec 2 at 1:37
Thomas Shelby
1,358216
ah ok, but how is that a Bernoulli distribution?
– BeepBoop
Dec 2 at 3:13
1
Do you mean Binomial?
– Thomas Shelby
Dec 2 at 6:23
Recall the uniqueness of moment generating functions.
– Thomas Shelby
Dec 2 at 6:27
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
begin{align}
M_{sum_{i=1}^{n}X_i}(t) &= prod_{i=1}^{n}M_{X_i}(t)\
& = prod_{i=1}^{n}(pe^t + 1 -p)^{n_i}\
&=(pe^t + 1 -p)^{sum_{i=1}^{n} n_i}
end{align}
answered Dec 2 at 1:37
Thomas Shelby
1,358216
ah ok, but how is that a Bernoulli distribution?
– BeepBoop
Dec 2 at 3:13
1
Do you mean Binomial?
– Thomas Shelby
Dec 2 at 6:23
Recall the uniqueness of moment generating functions.
– Thomas Shelby
Dec 2 at 6:27
add a comment |
begin{align}
M_{sum_{i=1}^{n}X_i}(t) &= prod_{i=1}^{n}M_{X_i}(t)\
& = prod_{i=1}^{n}(pe^t + 1 -p)^{n_i}\
&=(pe^t + 1 -p)^{sum_{i=1}^{n} n_i}
end{align}
answered Dec 2 at 1:37
Thomas Shelby
1,358216
ah ok, but how is that a Bernoulli distribution?
– BeepBoop
Dec 2 at 3:13
1
Do you mean Binomial?
– Thomas Shelby
Dec 2 at 6:23
Recall the uniqueness of moment generating functions.
– Thomas Shelby
Dec 2 at 6:27
add a comment |
begin{align}
M_{sum_{i=1}^{n}X_i}(t) &= prod_{i=1}^{n}M_{X_i}(t)\
& = prod_{i=1}^{n}(pe^t + 1 -p)^{n_i}\
&=(pe^t + 1 -p)^{sum_{i=1}^{n} n_i}
end{align}
answered Dec 2 at 1:37
Thomas Shelby
1,358216
begin{align}
M_{sum_{i=1}^{n}X_i}(t) &= prod_{i=1}^{n}M_{X_i}(t)\
& = prod_{i=1}^{n}(pe^t + 1 -p)^{n_i}\
&=(pe^t + 1 -p)^{sum_{i=1}^{n} n_i}
end{align}
answered Dec 2 at 1:37
Thomas Shelby
1,358216
edited Dec 2 at 6:42
answered Dec 2 at 1:37
Thomas Shelby
1,358216
answered Dec 2 at 1:37
Thomas Shelby
1,358216
answered Dec 2 at 1:37
Thomas Shelby
1,358216
1,358216
ah ok, but how is that a Bernoulli distribution?
– BeepBoop
Dec 2 at 3:13
1
Do you mean Binomial?
– Thomas Shelby
Dec 2 at 6:23
Recall the uniqueness of moment generating functions.
– Thomas Shelby
Dec 2 at 6:27
add a comment |
ah ok, but how is that a Bernoulli distribution?
– BeepBoop
Dec 2 at 3:13
1
Do you mean Binomial?
– Thomas Shelby
Dec 2 at 6:23
Recall the uniqueness of moment generating functions.
– Thomas Shelby
Dec 2 at 6:27
ah ok, but how is that a Bernoulli distribution?
– BeepBoop
Dec 2 at 3:13
ah ok, but how is that a Bernoulli distribution?
– BeepBoop
Dec 2 at 3:13
1
1
Do you mean Binomial?
– Thomas Shelby
Dec 2 at 6:23
Do you mean Binomial?
– Thomas Shelby
Dec 2 at 6:23
Recall the uniqueness of moment generating functions.
– Thomas Shelby
Dec 2 at 6:27
Recall the uniqueness of moment generating functions.
– Thomas Shelby
Dec 2 at 6:27
add a comment |
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