Central limit theorem for sequence of Gamma-distributed random variables.
up vote
2
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Suppose that $X_ n sim text {Gamma} (nalpha , lambda)$ for all $n ge 1$, for fixed $alpha,lambda >0.$ Show that
$$frac {1} {sqrt n} left (X_n - frac {n alpha} {lambda} right ) implies Z ,$$
where $Z sim mathcal N (0,{sigma}^2)$ for some $sigma.$ Calculate $sigma.$
I have constructed a sequence random variables ${S_n }$ in the following way $:$
$S_1=X_1,S_2=X_2-X_1,S_3=X_3-X_2, cdots , S_n = X_n - X_{n-1} , cdots.$ Then I observed that $sumlimits_{k=1}^{n} S_k = X_n$ for all $n ge 1$. Also I observed that $Bbb E(S_n) = frac {alpha} {lambda}$ and $Bbb {Var} (S_n) = frac {alpha} {{lambda}^2}$ for all $n ge 1$. If moreover $S_n$'s can be shown to be independent then they are i.i.d. random variables with mean $frac {alpha} {lambda}$ and variance $frac {alpha} {{lambda}^2}$. Then by central limit theorem we can say that
$$frac {1} {sqrt n} left (X_n - frac {n alpha} {lambda} right ) implies Z$$ as $n rightarrow infty$ where $Z sim mathcal N (0,frac {alpha} {{lambda}^2})$. Hence ${sigma}^2 = frac {alpha} {{lambda}^2}$ i.e. $sigma =frac {sqrt alpha} {lambda}.$
But how do I prove that $S_i$'s are i.i.d. random variables keeping in mind the fact that $X_i$'s are independent random variables? Please help me in this regard.
Thank you very much.
probability-theory convergence weak-convergence central-limit-theorem gamma-distribution
asked Dec 1 at 7:49
Dbchatto67
49515
|
show 1 more comment
up vote
2
down vote
favorite
Suppose that $X_ n sim text {Gamma} (nalpha , lambda)$ for all $n ge 1$, for fixed $alpha,lambda >0.$ Show that
$$frac {1} {sqrt n} left (X_n - frac {n alpha} {lambda} right ) implies Z ,$$
where $Z sim mathcal N (0,{sigma}^2)$ for some $sigma.$ Calculate $sigma.$
I have constructed a sequence random variables ${S_n }$ in the following way $:$
$S_1=X_1,S_2=X_2-X_1,S_3=X_3-X_2, cdots , S_n = X_n - X_{n-1} , cdots.$ Then I observed that $sumlimits_{k=1}^{n} S_k = X_n$ for all $n ge 1$. Also I observed that $Bbb E(S_n) = frac {alpha} {lambda}$ and $Bbb {Var} (S_n) = frac {alpha} {{lambda}^2}$ for all $n ge 1$. If moreover $S_n$'s can be shown to be independent then they are i.i.d. random variables with mean $frac {alpha} {lambda}$ and variance $frac {alpha} {{lambda}^2}$. Then by central limit theorem we can say that
$$frac {1} {sqrt n} left (X_n - frac {n alpha} {lambda} right ) implies Z$$ as $n rightarrow infty$ where $Z sim mathcal N (0,frac {alpha} {{lambda}^2})$. Hence ${sigma}^2 = frac {alpha} {{lambda}^2}$ i.e. $sigma =frac {sqrt alpha} {lambda}.$
But how do I prove that $S_i$'s are i.i.d. random variables keeping in mind the fact that $X_i$'s are independent random variables? Please help me in this regard.
Thank you very much.
probability-theory convergence weak-convergence central-limit-theorem gamma-distribution
asked Dec 1 at 7:49
Dbchatto67
49515
Regarding your last paragraph: the $X_i$ are not i.i.d., as their distributions depend on $i$.
– Quaternion
Dec 1 at 8:29
Sorry I mean to say that they are independent. Thanks for pointing this out @Quaternion. I will edit my question soon.
– Dbchatto67
Dec 1 at 8:36
@Dbchatto67 I don't think your variance of $S_n$ is calculated correctly. Note that $textrm{V}(S_n) = textrm{V}(X_n-X_{n-1}) = textrm{V}(X_n)+textrm{V}(X_{n-1}) = frac{alpha n}{lambda^2} + frac{alpha (n-1)}{lambda^2}$
– Lundborg
Dec 1 at 9:06
In the statement of your problem you do not mention independence of the $X_i$...
– saz
Dec 1 at 9:20
I'm not sure if introducing the $S_n$'s and using the CLT is a good idea. Have you tried directly computing the limit while using the Stirling approximation for the $Gamma(alpha n)$ term?
– Quaternion
Dec 1 at 9:21
|
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose that $X_ n sim text {Gamma} (nalpha , lambda)$ for all $n ge 1$, for fixed $alpha,lambda >0.$ Show that
$$frac {1} {sqrt n} left (X_n - frac {n alpha} {lambda} right ) implies Z ,$$
where $Z sim mathcal N (0,{sigma}^2)$ for some $sigma.$ Calculate $sigma.$
I have constructed a sequence random variables ${S_n }$ in the following way $:$
$S_1=X_1,S_2=X_2-X_1,S_3=X_3-X_2, cdots , S_n = X_n - X_{n-1} , cdots.$ Then I observed that $sumlimits_{k=1}^{n} S_k = X_n$ for all $n ge 1$. Also I observed that $Bbb E(S_n) = frac {alpha} {lambda}$ and $Bbb {Var} (S_n) = frac {alpha} {{lambda}^2}$ for all $n ge 1$. If moreover $S_n$'s can be shown to be independent then they are i.i.d. random variables with mean $frac {alpha} {lambda}$ and variance $frac {alpha} {{lambda}^2}$. Then by central limit theorem we can say that
$$frac {1} {sqrt n} left (X_n - frac {n alpha} {lambda} right ) implies Z$$ as $n rightarrow infty$ where $Z sim mathcal N (0,frac {alpha} {{lambda}^2})$. Hence ${sigma}^2 = frac {alpha} {{lambda}^2}$ i.e. $sigma =frac {sqrt alpha} {lambda}.$
But how do I prove that $S_i$'s are i.i.d. random variables keeping in mind the fact that $X_i$'s are independent random variables? Please help me in this regard.
Thank you very much.
probability-theory convergence weak-convergence central-limit-theorem gamma-distribution
asked Dec 1 at 7:49
Dbchatto67
49515
Suppose that $X_ n sim text {Gamma} (nalpha , lambda)$ for all $n ge 1$, for fixed $alpha,lambda >0.$ Show that
$$frac {1} {sqrt n} left (X_n - frac {n alpha} {lambda} right ) implies Z ,$$
where $Z sim mathcal N (0,{sigma}^2)$ for some $sigma.$ Calculate $sigma.$
I have constructed a sequence random variables ${S_n }$ in the following way $:$
$S_1=X_1,S_2=X_2-X_1,S_3=X_3-X_2, cdots , S_n = X_n - X_{n-1} , cdots.$ Then I observed that $sumlimits_{k=1}^{n} S_k = X_n$ for all $n ge 1$. Also I observed that $Bbb E(S_n) = frac {alpha} {lambda}$ and $Bbb {Var} (S_n) = frac {alpha} {{lambda}^2}$ for all $n ge 1$. If moreover $S_n$'s can be shown to be independent then they are i.i.d. random variables with mean $frac {alpha} {lambda}$ and variance $frac {alpha} {{lambda}^2}$. Then by central limit theorem we can say that
$$frac {1} {sqrt n} left (X_n - frac {n alpha} {lambda} right ) implies Z$$ as $n rightarrow infty$ where $Z sim mathcal N (0,frac {alpha} {{lambda}^2})$. Hence ${sigma}^2 = frac {alpha} {{lambda}^2}$ i.e. $sigma =frac {sqrt alpha} {lambda}.$
But how do I prove that $S_i$'s are i.i.d. random variables keeping in mind the fact that $X_i$'s are independent random variables? Please help me in this regard.
Thank you very much.
probability-theory convergence weak-convergence central-limit-theorem gamma-distribution
probability-theory convergence weak-convergence central-limit-theorem gamma-distribution
asked Dec 1 at 7:49
Dbchatto67
49515
asked Dec 1 at 7:49
Dbchatto67
49515
edited Dec 1 at 14:37
asked Dec 1 at 7:49
Dbchatto67
49515
asked Dec 1 at 7:49
Dbchatto67
49515
asked Dec 1 at 7:49
Dbchatto67
49515
49515
Regarding your last paragraph: the $X_i$ are not i.i.d., as their distributions depend on $i$.
– Quaternion
Dec 1 at 8:29
Sorry I mean to say that they are independent. Thanks for pointing this out @Quaternion. I will edit my question soon.
– Dbchatto67
Dec 1 at 8:36
@Dbchatto67 I don't think your variance of $S_n$ is calculated correctly. Note that $textrm{V}(S_n) = textrm{V}(X_n-X_{n-1}) = textrm{V}(X_n)+textrm{V}(X_{n-1}) = frac{alpha n}{lambda^2} + frac{alpha (n-1)}{lambda^2}$
– Lundborg
Dec 1 at 9:06
In the statement of your problem you do not mention independence of the $X_i$...
– saz
Dec 1 at 9:20
I'm not sure if introducing the $S_n$'s and using the CLT is a good idea. Have you tried directly computing the limit while using the Stirling approximation for the $Gamma(alpha n)$ term?
– Quaternion
Dec 1 at 9:21
|
show 1 more comment
Regarding your last paragraph: the $X_i$ are not i.i.d., as their distributions depend on $i$.
– Quaternion
Dec 1 at 8:29
Sorry I mean to say that they are independent. Thanks for pointing this out @Quaternion. I will edit my question soon.
– Dbchatto67
Dec 1 at 8:36
@Dbchatto67 I don't think your variance of $S_n$ is calculated correctly. Note that $textrm{V}(S_n) = textrm{V}(X_n-X_{n-1}) = textrm{V}(X_n)+textrm{V}(X_{n-1}) = frac{alpha n}{lambda^2} + frac{alpha (n-1)}{lambda^2}$
– Lundborg
Dec 1 at 9:06
In the statement of your problem you do not mention independence of the $X_i$...
– saz
Dec 1 at 9:20
I'm not sure if introducing the $S_n$'s and using the CLT is a good idea. Have you tried directly computing the limit while using the Stirling approximation for the $Gamma(alpha n)$ term?
– Quaternion
Dec 1 at 9:21
Regarding your last paragraph: the $X_i$ are not i.i.d., as their distributions depend on $i$.
– Quaternion
Dec 1 at 8:29
Regarding your last paragraph: the $X_i$ are not i.i.d., as their distributions depend on $i$.
– Quaternion
Dec 1 at 8:29
Sorry I mean to say that they are independent. Thanks for pointing this out @Quaternion. I will edit my question soon.
– Dbchatto67
Dec 1 at 8:36
Sorry I mean to say that they are independent. Thanks for pointing this out @Quaternion. I will edit my question soon.
– Dbchatto67
Dec 1 at 8:36
@Dbchatto67 I don't think your variance of $S_n$ is calculated correctly. Note that $textrm{V}(S_n) = textrm{V}(X_n-X_{n-1}) = textrm{V}(X_n)+textrm{V}(X_{n-1}) = frac{alpha n}{lambda^2} + frac{alpha (n-1)}{lambda^2}$
– Lundborg
Dec 1 at 9:06
@Dbchatto67 I don't think your variance of $S_n$ is calculated correctly. Note that $textrm{V}(S_n) = textrm{V}(X_n-X_{n-1}) = textrm{V}(X_n)+textrm{V}(X_{n-1}) = frac{alpha n}{lambda^2} + frac{alpha (n-1)}{lambda^2}$
– Lundborg
Dec 1 at 9:06
In the statement of your problem you do not mention independence of the $X_i$...
– saz
Dec 1 at 9:20
In the statement of your problem you do not mention independence of the $X_i$...
– saz
Dec 1 at 9:20
I'm not sure if introducing the $S_n$'s and using the CLT is a good idea. Have you tried directly computing the limit while using the Stirling approximation for the $Gamma(alpha n)$ term?
– Quaternion
Dec 1 at 9:21
I'm not sure if introducing the $S_n$'s and using the CLT is a good idea. Have you tried directly computing the limit while using the Stirling approximation for the $Gamma(alpha n)$ term?
– Quaternion
Dec 1 at 9:21
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
This answer uses the the following fact.
If $X sim Gamma(alpha,lambda)$ and $Y sim Gamma(beta,lambda)$ are independent, then $X+Y sim Gamma(alpha+beta,lambda)$.
Hints: Let $(Y_i)_{i in mathbb{N}}$ be a sequence of independent identically distributed random variables such that $Y_i sim Gamma(alpha,lambda)$.
- Show that $tilde{X}_n := sum_{i=1}^n Y_i$ satisfies $tilde{X}_n sim Gamma(n alpha,lambda)$.
- Apply the central limit theorem to prove that $$frac{1}{sqrt{n}} left( tilde{X}_n - frac{n alpha}{lambda} right) stackrel{d}{to} Z$$ for $Z sim N(0,sigma^2)$ with $sigma^2 = text{var}(Y_1)$; here $stackrel{d}{to}$ denotes convergence in distribution.
- Use the fact that $tilde{X}_n$ equals in distribution $X_n$ for each $n in mathbb{N}$ to conclude from Step 2 that $$frac{1}{sqrt{n}} left( X_n - frac{n alpha}{lambda} right) stackrel{d}{to} Z$$
- Compute $sigma^2 = text{var}(Y_1)$ (...or look it up, e.g. on wikipedia).
answered Dec 1 at 9:36
saz
77.7k755120
So $Z sim mathcal N left (0, frac {alpha} {{lambda}^2} right )$. Then $sigma = frac {sqrt alpha} {lambda}$. Am I right @saz?
– Dbchatto67
Dec 1 at 11:28
@Dbchatto67 Yes, that's right.
– saz
Dec 1 at 11:50
Could you take a look at my last question? would really appreciate it
– orange
Dec 1 at 14:10
@itry To which end? There is an answer and you accepted it... so where is the problem?
– saz
Dec 1 at 17:23
I mean the one with the bounty about cramers theorem
– orange
Dec 1 at 17:24
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
up vote
6
down vote
accepted
This answer uses the the following fact.
If $X sim Gamma(alpha,lambda)$ and $Y sim Gamma(beta,lambda)$ are independent, then $X+Y sim Gamma(alpha+beta,lambda)$.
Hints: Let $(Y_i)_{i in mathbb{N}}$ be a sequence of independent identically distributed random variables such that $Y_i sim Gamma(alpha,lambda)$.
- Show that $tilde{X}_n := sum_{i=1}^n Y_i$ satisfies $tilde{X}_n sim Gamma(n alpha,lambda)$.
- Apply the central limit theorem to prove that $$frac{1}{sqrt{n}} left( tilde{X}_n - frac{n alpha}{lambda} right) stackrel{d}{to} Z$$ for $Z sim N(0,sigma^2)$ with $sigma^2 = text{var}(Y_1)$; here $stackrel{d}{to}$ denotes convergence in distribution.
- Use the fact that $tilde{X}_n$ equals in distribution $X_n$ for each $n in mathbb{N}$ to conclude from Step 2 that $$frac{1}{sqrt{n}} left( X_n - frac{n alpha}{lambda} right) stackrel{d}{to} Z$$
- Compute $sigma^2 = text{var}(Y_1)$ (...or look it up, e.g. on wikipedia).
answered Dec 1 at 9:36
saz
77.7k755120
So $Z sim mathcal N left (0, frac {alpha} {{lambda}^2} right )$. Then $sigma = frac {sqrt alpha} {lambda}$. Am I right @saz?
– Dbchatto67
Dec 1 at 11:28
@Dbchatto67 Yes, that's right.
– saz
Dec 1 at 11:50
Could you take a look at my last question? would really appreciate it
– orange
Dec 1 at 14:10
@itry To which end? There is an answer and you accepted it... so where is the problem?
– saz
Dec 1 at 17:23
I mean the one with the bounty about cramers theorem
– orange
Dec 1 at 17:24
add a comment |
up vote
6
down vote
accepted
This answer uses the the following fact.
If $X sim Gamma(alpha,lambda)$ and $Y sim Gamma(beta,lambda)$ are independent, then $X+Y sim Gamma(alpha+beta,lambda)$.
Hints: Let $(Y_i)_{i in mathbb{N}}$ be a sequence of independent identically distributed random variables such that $Y_i sim Gamma(alpha,lambda)$.
- Show that $tilde{X}_n := sum_{i=1}^n Y_i$ satisfies $tilde{X}_n sim Gamma(n alpha,lambda)$.
- Apply the central limit theorem to prove that $$frac{1}{sqrt{n}} left( tilde{X}_n - frac{n alpha}{lambda} right) stackrel{d}{to} Z$$ for $Z sim N(0,sigma^2)$ with $sigma^2 = text{var}(Y_1)$; here $stackrel{d}{to}$ denotes convergence in distribution.
- Use the fact that $tilde{X}_n$ equals in distribution $X_n$ for each $n in mathbb{N}$ to conclude from Step 2 that $$frac{1}{sqrt{n}} left( X_n - frac{n alpha}{lambda} right) stackrel{d}{to} Z$$
- Compute $sigma^2 = text{var}(Y_1)$ (...or look it up, e.g. on wikipedia).
answered Dec 1 at 9:36
saz
77.7k755120
So $Z sim mathcal N left (0, frac {alpha} {{lambda}^2} right )$. Then $sigma = frac {sqrt alpha} {lambda}$. Am I right @saz?
– Dbchatto67
Dec 1 at 11:28
@Dbchatto67 Yes, that's right.
– saz
Dec 1 at 11:50
Could you take a look at my last question? would really appreciate it
– orange
Dec 1 at 14:10
@itry To which end? There is an answer and you accepted it... so where is the problem?
– saz
Dec 1 at 17:23
I mean the one with the bounty about cramers theorem
– orange
Dec 1 at 17:24
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
This answer uses the the following fact.
If $X sim Gamma(alpha,lambda)$ and $Y sim Gamma(beta,lambda)$ are independent, then $X+Y sim Gamma(alpha+beta,lambda)$.
Hints: Let $(Y_i)_{i in mathbb{N}}$ be a sequence of independent identically distributed random variables such that $Y_i sim Gamma(alpha,lambda)$.
- Show that $tilde{X}_n := sum_{i=1}^n Y_i$ satisfies $tilde{X}_n sim Gamma(n alpha,lambda)$.
- Apply the central limit theorem to prove that $$frac{1}{sqrt{n}} left( tilde{X}_n - frac{n alpha}{lambda} right) stackrel{d}{to} Z$$ for $Z sim N(0,sigma^2)$ with $sigma^2 = text{var}(Y_1)$; here $stackrel{d}{to}$ denotes convergence in distribution.
- Use the fact that $tilde{X}_n$ equals in distribution $X_n$ for each $n in mathbb{N}$ to conclude from Step 2 that $$frac{1}{sqrt{n}} left( X_n - frac{n alpha}{lambda} right) stackrel{d}{to} Z$$
- Compute $sigma^2 = text{var}(Y_1)$ (...or look it up, e.g. on wikipedia).
answered Dec 1 at 9:36
saz
77.7k755120
This answer uses the the following fact.
If $X sim Gamma(alpha,lambda)$ and $Y sim Gamma(beta,lambda)$ are independent, then $X+Y sim Gamma(alpha+beta,lambda)$.
Hints: Let $(Y_i)_{i in mathbb{N}}$ be a sequence of independent identically distributed random variables such that $Y_i sim Gamma(alpha,lambda)$.
- Show that $tilde{X}_n := sum_{i=1}^n Y_i$ satisfies $tilde{X}_n sim Gamma(n alpha,lambda)$.
- Apply the central limit theorem to prove that $$frac{1}{sqrt{n}} left( tilde{X}_n - frac{n alpha}{lambda} right) stackrel{d}{to} Z$$ for $Z sim N(0,sigma^2)$ with $sigma^2 = text{var}(Y_1)$; here $stackrel{d}{to}$ denotes convergence in distribution.
- Use the fact that $tilde{X}_n$ equals in distribution $X_n$ for each $n in mathbb{N}$ to conclude from Step 2 that $$frac{1}{sqrt{n}} left( X_n - frac{n alpha}{lambda} right) stackrel{d}{to} Z$$
- Compute $sigma^2 = text{var}(Y_1)$ (...or look it up, e.g. on wikipedia).
answered Dec 1 at 9:36
saz
77.7k755120
answered Dec 1 at 9:36
saz
77.7k755120
answered Dec 1 at 9:36
saz
77.7k755120
answered Dec 1 at 9:36
saz
77.7k755120
77.7k755120
So $Z sim mathcal N left (0, frac {alpha} {{lambda}^2} right )$. Then $sigma = frac {sqrt alpha} {lambda}$. Am I right @saz?
– Dbchatto67
Dec 1 at 11:28
@Dbchatto67 Yes, that's right.
– saz
Dec 1 at 11:50
Could you take a look at my last question? would really appreciate it
– orange
Dec 1 at 14:10
@itry To which end? There is an answer and you accepted it... so where is the problem?
– saz
Dec 1 at 17:23
I mean the one with the bounty about cramers theorem
– orange
Dec 1 at 17:24
add a comment |
So $Z sim mathcal N left (0, frac {alpha} {{lambda}^2} right )$. Then $sigma = frac {sqrt alpha} {lambda}$. Am I right @saz?
– Dbchatto67
Dec 1 at 11:28
@Dbchatto67 Yes, that's right.
– saz
Dec 1 at 11:50
Could you take a look at my last question? would really appreciate it
– orange
Dec 1 at 14:10
@itry To which end? There is an answer and you accepted it... so where is the problem?
– saz
Dec 1 at 17:23
I mean the one with the bounty about cramers theorem
– orange
Dec 1 at 17:24
So $Z sim mathcal N left (0, frac {alpha} {{lambda}^2} right )$. Then $sigma = frac {sqrt alpha} {lambda}$. Am I right @saz?
– Dbchatto67
Dec 1 at 11:28
So $Z sim mathcal N left (0, frac {alpha} {{lambda}^2} right )$. Then $sigma = frac {sqrt alpha} {lambda}$. Am I right @saz?
– Dbchatto67
Dec 1 at 11:28
@Dbchatto67 Yes, that's right.
– saz
Dec 1 at 11:50
@Dbchatto67 Yes, that's right.
– saz
Dec 1 at 11:50
Could you take a look at my last question? would really appreciate it
– orange
Dec 1 at 14:10
Could you take a look at my last question? would really appreciate it
– orange
Dec 1 at 14:10
@itry To which end? There is an answer and you accepted it... so where is the problem?
– saz
Dec 1 at 17:23
@itry To which end? There is an answer and you accepted it... so where is the problem?
– saz
Dec 1 at 17:23
I mean the one with the bounty about cramers theorem
– orange
Dec 1 at 17:24
I mean the one with the bounty about cramers theorem
– orange
Dec 1 at 17:24
add a comment |
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Regarding your last paragraph: the $X_i$ are not i.i.d., as their distributions depend on $i$.
– Quaternion
Dec 1 at 8:29
Sorry I mean to say that they are independent. Thanks for pointing this out @Quaternion. I will edit my question soon.
– Dbchatto67
Dec 1 at 8:36
@Dbchatto67 I don't think your variance of $S_n$ is calculated correctly. Note that $textrm{V}(S_n) = textrm{V}(X_n-X_{n-1}) = textrm{V}(X_n)+textrm{V}(X_{n-1}) = frac{alpha n}{lambda^2} + frac{alpha (n-1)}{lambda^2}$
– Lundborg
Dec 1 at 9:06
In the statement of your problem you do not mention independence of the $X_i$...
– saz
Dec 1 at 9:20
I'm not sure if introducing the $S_n$'s and using the CLT is a good idea. Have you tried directly computing the limit while using the Stirling approximation for the $Gamma(alpha n)$ term?
– Quaternion
Dec 1 at 9:21