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Let $X sim Bin(n,p)$ show that $X/n nsim Bin()$
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Let $X sim Bin(n,p)$
Part A.
Show that the argument
"Then $X/n sim Bin()$ with $E[X/n]=p$and $Var[X/n]=pq/n$"
is false.
My book says we can prove this result using a moment generating function. But to me it makes no intuitive sense: We have Bin-distributed RV $X$ and if we rescale it with $1/n$ of course it will be Bin-distributed.
And we know from standard formulas that $E[aX]=aE[X]$ and $Var[aX] = a^2Var[X]$ so I think the argument is correct.
(For full disclosure, the argument we should falsify according to the book is "Let X be a binomial random variable with mean np and variance np(1−p). Show that the ratio X/n also has a binomial distribution with mean p and variance p(1−p)/n." But I think the argument is correct. )
Ok, now I get part A. Here is Part B. Prove it using the moment generating function of $X/n$.
Attempt: I get $$m_X(t)=E[exp(tX/n)]=(m_X(t))^1/n$$
probability-theory
asked Sep 16 '15 at 21:52
jacob
1,21221226
add a comment |Â
up vote
0
down vote
favorite
Let $X sim Bin(n,p)$
Part A.
Show that the argument
"Then $X/n sim Bin()$ with $E[X/n]=p$and $Var[X/n]=pq/n$"
is false.
My book says we can prove this result using a moment generating function. But to me it makes no intuitive sense: We have Bin-distributed RV $X$ and if we rescale it with $1/n$ of course it will be Bin-distributed.
And we know from standard formulas that $E[aX]=aE[X]$ and $Var[aX] = a^2Var[X]$ so I think the argument is correct.
(For full disclosure, the argument we should falsify according to the book is "Let X be a binomial random variable with mean np and variance np(1−p). Show that the ratio X/n also has a binomial distribution with mean p and variance p(1−p)/n." But I think the argument is correct. )
Ok, now I get part A. Here is Part B. Prove it using the moment generating function of $X/n$.
Attempt: I get $$m_X(t)=E[exp(tX/n)]=(m_X(t))^1/n$$
probability-theory
asked Sep 16 '15 at 21:52
jacob
1,21221226
2
I binomially distributed random variable $X$ with $n=6$ has support $0,1,2,3,4,5,6$. But $X/6$ has support $0 ,, 1/6,, 2/6,, 3/6,, 4/6,, 5/6,, 1$. That is not the support of a binomial distribution. $qquad$
– Michael Hardy
Sep 16 '15 at 22:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X sim Bin(n,p)$
Part A.
Show that the argument
"Then $X/n sim Bin()$ with $E[X/n]=p$and $Var[X/n]=pq/n$"
is false.
My book says we can prove this result using a moment generating function. But to me it makes no intuitive sense: We have Bin-distributed RV $X$ and if we rescale it with $1/n$ of course it will be Bin-distributed.
And we know from standard formulas that $E[aX]=aE[X]$ and $Var[aX] = a^2Var[X]$ so I think the argument is correct.
(For full disclosure, the argument we should falsify according to the book is "Let X be a binomial random variable with mean np and variance np(1−p). Show that the ratio X/n also has a binomial distribution with mean p and variance p(1−p)/n." But I think the argument is correct. )
Ok, now I get part A. Here is Part B. Prove it using the moment generating function of $X/n$.
Attempt: I get $$m_X(t)=E[exp(tX/n)]=(m_X(t))^1/n$$
probability-theory
asked Sep 16 '15 at 21:52
jacob
1,21221226
Let $X sim Bin(n,p)$
Part A.
Show that the argument
"Then $X/n sim Bin()$ with $E[X/n]=p$and $Var[X/n]=pq/n$"
is false.
My book says we can prove this result using a moment generating function. But to me it makes no intuitive sense: We have Bin-distributed RV $X$ and if we rescale it with $1/n$ of course it will be Bin-distributed.
And we know from standard formulas that $E[aX]=aE[X]$ and $Var[aX] = a^2Var[X]$ so I think the argument is correct.
(For full disclosure, the argument we should falsify according to the book is "Let X be a binomial random variable with mean np and variance np(1−p). Show that the ratio X/n also has a binomial distribution with mean p and variance p(1−p)/n." But I think the argument is correct. )
Ok, now I get part A. Here is Part B. Prove it using the moment generating function of $X/n$.
Attempt: I get $$m_X(t)=E[exp(tX/n)]=(m_X(t))^1/n$$
probability-theory
asked Sep 16 '15 at 21:52
jacob
1,21221226
edited Sep 17 '15 at 22:39
asked Sep 16 '15 at 21:52
jacob
1,21221226
asked Sep 16 '15 at 21:52
jacob
1,21221226
asked Sep 16 '15 at 21:52
jacob
1,21221226
1,21221226
2
I binomially distributed random variable $X$ with $n=6$ has support $0,1,2,3,4,5,6$. But $X/6$ has support $0 ,, 1/6,, 2/6,, 3/6,, 4/6,, 5/6,, 1$. That is not the support of a binomial distribution. $qquad$
– Michael Hardy
Sep 16 '15 at 22:12
add a comment |Â
2
I binomially distributed random variable $X$ with $n=6$ has support $0,1,2,3,4,5,6$. But $X/6$ has support $0 ,, 1/6,, 2/6,, 3/6,, 4/6,, 5/6,, 1$. That is not the support of a binomial distribution. $qquad$
– Michael Hardy
Sep 16 '15 at 22:12
2
2
I binomially distributed random variable $X$ with $n=6$ has support $0,1,2,3,4,5,6$. But $X/6$ has support $0 ,, 1/6,, 2/6,, 3/6,, 4/6,, 5/6,, 1$. That is not the support of a binomial distribution. $qquad$
– Michael Hardy
Sep 16 '15 at 22:12
I binomially distributed random variable $X$ with $n=6$ has support $0,1,2,3,4,5,6$. But $X/6$ has support $0 ,, 1/6,, 2/6,, 3/6,, 4/6,, 5/6,, 1$. That is not the support of a binomial distribution. $qquad$
– Michael Hardy
Sep 16 '15 at 22:12
add a comment |Â
1 Answer
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3
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Of course $fracXn$ isn't binomial distributed for $n > 1$, since $P(fracXn = frac1n) ne 0$ and the binomial distribution only assumes integer values.
You argument shows nothing. Just because a random variable has a mean and a variance doesn't mean it is binomial distributed.
answered Sep 16 '15 at 21:58
Dominik
17.4k11844
Thank you, this solved it. But I forgot one part of the question: Prove it using the moment generating function of $X/n$ and I get $$m_X(t)=E[exp(tX/n)]=(m_X(t))^1/n$$
– jacob
Sep 17 '15 at 22:38
You can't just pull $1/n$ out of the expectation. The correct result is $m_X(t/n)$.
– Dominik
Sep 18 '15 at 7:28
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Of course $fracXn$ isn't binomial distributed for $n > 1$, since $P(fracXn = frac1n) ne 0$ and the binomial distribution only assumes integer values.
You argument shows nothing. Just because a random variable has a mean and a variance doesn't mean it is binomial distributed.
answered Sep 16 '15 at 21:58
Dominik
17.4k11844
Thank you, this solved it. But I forgot one part of the question: Prove it using the moment generating function of $X/n$ and I get $$m_X(t)=E[exp(tX/n)]=(m_X(t))^1/n$$
– jacob
Sep 17 '15 at 22:38
You can't just pull $1/n$ out of the expectation. The correct result is $m_X(t/n)$.
– Dominik
Sep 18 '15 at 7:28
add a comment |Â
up vote
3
down vote
accepted
Of course $fracXn$ isn't binomial distributed for $n > 1$, since $P(fracXn = frac1n) ne 0$ and the binomial distribution only assumes integer values.
You argument shows nothing. Just because a random variable has a mean and a variance doesn't mean it is binomial distributed.
answered Sep 16 '15 at 21:58
Dominik
17.4k11844
Thank you, this solved it. But I forgot one part of the question: Prove it using the moment generating function of $X/n$ and I get $$m_X(t)=E[exp(tX/n)]=(m_X(t))^1/n$$
– jacob
Sep 17 '15 at 22:38
You can't just pull $1/n$ out of the expectation. The correct result is $m_X(t/n)$.
– Dominik
Sep 18 '15 at 7:28
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Of course $fracXn$ isn't binomial distributed for $n > 1$, since $P(fracXn = frac1n) ne 0$ and the binomial distribution only assumes integer values.
You argument shows nothing. Just because a random variable has a mean and a variance doesn't mean it is binomial distributed.
answered Sep 16 '15 at 21:58
Dominik
17.4k11844
Of course $fracXn$ isn't binomial distributed for $n > 1$, since $P(fracXn = frac1n) ne 0$ and the binomial distribution only assumes integer values.
You argument shows nothing. Just because a random variable has a mean and a variance doesn't mean it is binomial distributed.
answered Sep 16 '15 at 21:58
Dominik
17.4k11844
answered Sep 16 '15 at 21:58
Dominik
17.4k11844
answered Sep 16 '15 at 21:58
Dominik
17.4k11844
answered Sep 16 '15 at 21:58
Dominik
17.4k11844
17.4k11844
Thank you, this solved it. But I forgot one part of the question: Prove it using the moment generating function of $X/n$ and I get $$m_X(t)=E[exp(tX/n)]=(m_X(t))^1/n$$
– jacob
Sep 17 '15 at 22:38
You can't just pull $1/n$ out of the expectation. The correct result is $m_X(t/n)$.
– Dominik
Sep 18 '15 at 7:28
add a comment |Â
Thank you, this solved it. But I forgot one part of the question: Prove it using the moment generating function of $X/n$ and I get $$m_X(t)=E[exp(tX/n)]=(m_X(t))^1/n$$
– jacob
Sep 17 '15 at 22:38
You can't just pull $1/n$ out of the expectation. The correct result is $m_X(t/n)$.
– Dominik
Sep 18 '15 at 7:28
Thank you, this solved it. But I forgot one part of the question: Prove it using the moment generating function of $X/n$ and I get $$m_X(t)=E[exp(tX/n)]=(m_X(t))^1/n$$
– jacob
Sep 17 '15 at 22:38
Thank you, this solved it. But I forgot one part of the question: Prove it using the moment generating function of $X/n$ and I get $$m_X(t)=E[exp(tX/n)]=(m_X(t))^1/n$$
– jacob
Sep 17 '15 at 22:38
You can't just pull $1/n$ out of the expectation. The correct result is $m_X(t/n)$.
– Dominik
Sep 18 '15 at 7:28
You can't just pull $1/n$ out of the expectation. The correct result is $m_X(t/n)$.
– Dominik
Sep 18 '15 at 7:28
add a comment |Â
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2
I binomially distributed random variable $X$ with $n=6$ has support $0,1,2,3,4,5,6$. But $X/6$ has support $0 ,, 1/6,, 2/6,, 3/6,, 4/6,, 5/6,, 1$. That is not the support of a binomial distribution. $qquad$
– Michael Hardy
Sep 16 '15 at 22:12