Probability limits of random variable sums
$begingroup$
I have $X_1, X_2, X_3, cdots$ which are independent random variables with the same non-zero mean ($mune0$) and same variance $sigma^2$.
I would like to compute $$lim_{ntoinfty} P[frac{1}nsum^n_{i=1}X_i < frac{mu}{2}]$$ for $mu<0$ and $mu>0$.
My initial thought was to use the central limit theorem but it indicates the variables must be identically distributed which I dont have here -- only have first and second order moments are similar.
Any thoughts on how to start tackling this?
convergence central-limit-theorem probability-limit-theorems
$endgroup$
add a comment |
$begingroup$
I have $X_1, X_2, X_3, cdots$ which are independent random variables with the same non-zero mean ($mune0$) and same variance $sigma^2$.
I would like to compute $$lim_{ntoinfty} P[frac{1}nsum^n_{i=1}X_i < frac{mu}{2}]$$ for $mu<0$ and $mu>0$.
My initial thought was to use the central limit theorem but it indicates the variables must be identically distributed which I dont have here -- only have first and second order moments are similar.
Any thoughts on how to start tackling this?
convergence central-limit-theorem probability-limit-theorems
$endgroup$
add a comment |
$begingroup$
I have $X_1, X_2, X_3, cdots$ which are independent random variables with the same non-zero mean ($mune0$) and same variance $sigma^2$.
I would like to compute $$lim_{ntoinfty} P[frac{1}nsum^n_{i=1}X_i < frac{mu}{2}]$$ for $mu<0$ and $mu>0$.
My initial thought was to use the central limit theorem but it indicates the variables must be identically distributed which I dont have here -- only have first and second order moments are similar.
Any thoughts on how to start tackling this?
convergence central-limit-theorem probability-limit-theorems
$endgroup$
I have $X_1, X_2, X_3, cdots$ which are independent random variables with the same non-zero mean ($mune0$) and same variance $sigma^2$.
I would like to compute $$lim_{ntoinfty} P[frac{1}nsum^n_{i=1}X_i < frac{mu}{2}]$$ for $mu<0$ and $mu>0$.
My initial thought was to use the central limit theorem but it indicates the variables must be identically distributed which I dont have here -- only have first and second order moments are similar.
Any thoughts on how to start tackling this?
convergence central-limit-theorem probability-limit-theorems
convergence central-limit-theorem probability-limit-theorems
asked Dec 21 '18 at 2:28
AvedisAvedis
667
667
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Chebychev's inequality may help.
For $mu > 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| > frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0.$$ For $mu < 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i ge frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| ge frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0$$ so $Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) to 1$.
$endgroup$
1
$begingroup$
@ClementC. You're right, thanks for catching my mistake.
$endgroup$
– angryavian
Dec 21 '18 at 2:56
$begingroup$
Why is the right side of Chebychev's inquality have n and not n^2?
$endgroup$
– Avedis
Dec 23 '18 at 1:43
$begingroup$
@Avedis Think carefully about what the variance of $frac{1}{n} sum_{i=1}^n X_i$ is.
$endgroup$
– angryavian
Dec 23 '18 at 2:08
$begingroup$
Thank you, I just figure out what you mentioned on my own. However, when I complete the problem, I get a different answer then you do for $mu>0$. Why?
$endgroup$
– Avedis
Dec 23 '18 at 3:51
add a comment |
$begingroup$
Define $M_n=frac{1}{n}sum_{i-1}^nX_i$.
$E[M_n]=mu$, $Var[M_n]=frac{sigma_X^2}{n}$
For $mu<0$:
$$P[M_n<frac{mu}{2}]$$
$$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)
$$P[M_n-mu<frac{mu}{2}-mu]$$
$$P[M_n-mu< -frac{mu}{2}]$$
(using Chebyshev inequality)
$$P[|M_n-mu| < -frac{mu}{2}]$$
$$P[|M_n-mu| < -frac{mu}{2}] geq 1 - frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity, so the Probability is $1-0=1$.
For $mu > 0$:
$$P[M_n<frac{mu}{2}]$$
$$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)
$$P[M_n-mu<frac{mu}{2}-mu]$$
$$P[M_n-mu< -frac{mu}{2}]$$
(flip inequality due to division of both sides by -1)
$$P[-M_n+mu > frac{mu}{2}]$$
(using Chebyshev inequality and fact of $|A|=|-A|$)
$$P[|M_n-mu| > frac{mu}{2}]$$
$$P[|M_n-mu| < frac{mu}{2}] leq frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity. So, the probability is 0.
$endgroup$
$begingroup$
In both cases you should have $(mu/2)^2=mu^2/4$ rather than $mu^2/2$. And it seems our answers agree.
$endgroup$
– angryavian
Dec 23 '18 at 4:20
$begingroup$
You are right and I will fix. I misread your answer. I thought the last statement was a summary of the entire problem, not the last part exclusively. Thank you!
$endgroup$
– Avedis
Dec 23 '18 at 14:04
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
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$begingroup$
Chebychev's inequality may help.
For $mu > 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| > frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0.$$ For $mu < 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i ge frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| ge frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0$$ so $Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) to 1$.
$endgroup$
1
$begingroup$
@ClementC. You're right, thanks for catching my mistake.
$endgroup$
– angryavian
Dec 21 '18 at 2:56
$begingroup$
Why is the right side of Chebychev's inquality have n and not n^2?
$endgroup$
– Avedis
Dec 23 '18 at 1:43
$begingroup$
@Avedis Think carefully about what the variance of $frac{1}{n} sum_{i=1}^n X_i$ is.
$endgroup$
– angryavian
Dec 23 '18 at 2:08
$begingroup$
Thank you, I just figure out what you mentioned on my own. However, when I complete the problem, I get a different answer then you do for $mu>0$. Why?
$endgroup$
– Avedis
Dec 23 '18 at 3:51
add a comment |
$begingroup$
Chebychev's inequality may help.
For $mu > 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| > frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0.$$ For $mu < 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i ge frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| ge frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0$$ so $Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) to 1$.
$endgroup$
1
$begingroup$
@ClementC. You're right, thanks for catching my mistake.
$endgroup$
– angryavian
Dec 21 '18 at 2:56
$begingroup$
Why is the right side of Chebychev's inquality have n and not n^2?
$endgroup$
– Avedis
Dec 23 '18 at 1:43
$begingroup$
@Avedis Think carefully about what the variance of $frac{1}{n} sum_{i=1}^n X_i$ is.
$endgroup$
– angryavian
Dec 23 '18 at 2:08
$begingroup$
Thank you, I just figure out what you mentioned on my own. However, when I complete the problem, I get a different answer then you do for $mu>0$. Why?
$endgroup$
– Avedis
Dec 23 '18 at 3:51
add a comment |
$begingroup$
Chebychev's inequality may help.
For $mu > 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| > frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0.$$ For $mu < 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i ge frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| ge frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0$$ so $Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) to 1$.
$endgroup$
Chebychev's inequality may help.
For $mu > 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| > frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0.$$ For $mu < 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i ge frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| ge frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0$$ so $Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) to 1$.
edited Dec 21 '18 at 2:56
answered Dec 21 '18 at 2:33
angryavianangryavian
42.6k23481
42.6k23481
1
$begingroup$
@ClementC. You're right, thanks for catching my mistake.
$endgroup$
– angryavian
Dec 21 '18 at 2:56
$begingroup$
Why is the right side of Chebychev's inquality have n and not n^2?
$endgroup$
– Avedis
Dec 23 '18 at 1:43
$begingroup$
@Avedis Think carefully about what the variance of $frac{1}{n} sum_{i=1}^n X_i$ is.
$endgroup$
– angryavian
Dec 23 '18 at 2:08
$begingroup$
Thank you, I just figure out what you mentioned on my own. However, when I complete the problem, I get a different answer then you do for $mu>0$. Why?
$endgroup$
– Avedis
Dec 23 '18 at 3:51
add a comment |
1
$begingroup$
@ClementC. You're right, thanks for catching my mistake.
$endgroup$
– angryavian
Dec 21 '18 at 2:56
$begingroup$
Why is the right side of Chebychev's inquality have n and not n^2?
$endgroup$
– Avedis
Dec 23 '18 at 1:43
$begingroup$
@Avedis Think carefully about what the variance of $frac{1}{n} sum_{i=1}^n X_i$ is.
$endgroup$
– angryavian
Dec 23 '18 at 2:08
$begingroup$
Thank you, I just figure out what you mentioned on my own. However, when I complete the problem, I get a different answer then you do for $mu>0$. Why?
$endgroup$
– Avedis
Dec 23 '18 at 3:51
1
1
$begingroup$
@ClementC. You're right, thanks for catching my mistake.
$endgroup$
– angryavian
Dec 21 '18 at 2:56
$begingroup$
@ClementC. You're right, thanks for catching my mistake.
$endgroup$
– angryavian
Dec 21 '18 at 2:56
$begingroup$
Why is the right side of Chebychev's inquality have n and not n^2?
$endgroup$
– Avedis
Dec 23 '18 at 1:43
$begingroup$
Why is the right side of Chebychev's inquality have n and not n^2?
$endgroup$
– Avedis
Dec 23 '18 at 1:43
$begingroup$
@Avedis Think carefully about what the variance of $frac{1}{n} sum_{i=1}^n X_i$ is.
$endgroup$
– angryavian
Dec 23 '18 at 2:08
$begingroup$
@Avedis Think carefully about what the variance of $frac{1}{n} sum_{i=1}^n X_i$ is.
$endgroup$
– angryavian
Dec 23 '18 at 2:08
$begingroup$
Thank you, I just figure out what you mentioned on my own. However, when I complete the problem, I get a different answer then you do for $mu>0$. Why?
$endgroup$
– Avedis
Dec 23 '18 at 3:51
$begingroup$
Thank you, I just figure out what you mentioned on my own. However, when I complete the problem, I get a different answer then you do for $mu>0$. Why?
$endgroup$
– Avedis
Dec 23 '18 at 3:51
add a comment |
$begingroup$
Define $M_n=frac{1}{n}sum_{i-1}^nX_i$.
$E[M_n]=mu$, $Var[M_n]=frac{sigma_X^2}{n}$
For $mu<0$:
$$P[M_n<frac{mu}{2}]$$
$$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)
$$P[M_n-mu<frac{mu}{2}-mu]$$
$$P[M_n-mu< -frac{mu}{2}]$$
(using Chebyshev inequality)
$$P[|M_n-mu| < -frac{mu}{2}]$$
$$P[|M_n-mu| < -frac{mu}{2}] geq 1 - frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity, so the Probability is $1-0=1$.
For $mu > 0$:
$$P[M_n<frac{mu}{2}]$$
$$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)
$$P[M_n-mu<frac{mu}{2}-mu]$$
$$P[M_n-mu< -frac{mu}{2}]$$
(flip inequality due to division of both sides by -1)
$$P[-M_n+mu > frac{mu}{2}]$$
(using Chebyshev inequality and fact of $|A|=|-A|$)
$$P[|M_n-mu| > frac{mu}{2}]$$
$$P[|M_n-mu| < frac{mu}{2}] leq frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity. So, the probability is 0.
$endgroup$
$begingroup$
In both cases you should have $(mu/2)^2=mu^2/4$ rather than $mu^2/2$. And it seems our answers agree.
$endgroup$
– angryavian
Dec 23 '18 at 4:20
$begingroup$
You are right and I will fix. I misread your answer. I thought the last statement was a summary of the entire problem, not the last part exclusively. Thank you!
$endgroup$
– Avedis
Dec 23 '18 at 14:04
add a comment |
$begingroup$
Define $M_n=frac{1}{n}sum_{i-1}^nX_i$.
$E[M_n]=mu$, $Var[M_n]=frac{sigma_X^2}{n}$
For $mu<0$:
$$P[M_n<frac{mu}{2}]$$
$$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)
$$P[M_n-mu<frac{mu}{2}-mu]$$
$$P[M_n-mu< -frac{mu}{2}]$$
(using Chebyshev inequality)
$$P[|M_n-mu| < -frac{mu}{2}]$$
$$P[|M_n-mu| < -frac{mu}{2}] geq 1 - frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity, so the Probability is $1-0=1$.
For $mu > 0$:
$$P[M_n<frac{mu}{2}]$$
$$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)
$$P[M_n-mu<frac{mu}{2}-mu]$$
$$P[M_n-mu< -frac{mu}{2}]$$
(flip inequality due to division of both sides by -1)
$$P[-M_n+mu > frac{mu}{2}]$$
(using Chebyshev inequality and fact of $|A|=|-A|$)
$$P[|M_n-mu| > frac{mu}{2}]$$
$$P[|M_n-mu| < frac{mu}{2}] leq frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity. So, the probability is 0.
$endgroup$
$begingroup$
In both cases you should have $(mu/2)^2=mu^2/4$ rather than $mu^2/2$. And it seems our answers agree.
$endgroup$
– angryavian
Dec 23 '18 at 4:20
$begingroup$
You are right and I will fix. I misread your answer. I thought the last statement was a summary of the entire problem, not the last part exclusively. Thank you!
$endgroup$
– Avedis
Dec 23 '18 at 14:04
add a comment |
$begingroup$
Define $M_n=frac{1}{n}sum_{i-1}^nX_i$.
$E[M_n]=mu$, $Var[M_n]=frac{sigma_X^2}{n}$
For $mu<0$:
$$P[M_n<frac{mu}{2}]$$
$$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)
$$P[M_n-mu<frac{mu}{2}-mu]$$
$$P[M_n-mu< -frac{mu}{2}]$$
(using Chebyshev inequality)
$$P[|M_n-mu| < -frac{mu}{2}]$$
$$P[|M_n-mu| < -frac{mu}{2}] geq 1 - frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity, so the Probability is $1-0=1$.
For $mu > 0$:
$$P[M_n<frac{mu}{2}]$$
$$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)
$$P[M_n-mu<frac{mu}{2}-mu]$$
$$P[M_n-mu< -frac{mu}{2}]$$
(flip inequality due to division of both sides by -1)
$$P[-M_n+mu > frac{mu}{2}]$$
(using Chebyshev inequality and fact of $|A|=|-A|$)
$$P[|M_n-mu| > frac{mu}{2}]$$
$$P[|M_n-mu| < frac{mu}{2}] leq frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity. So, the probability is 0.
$endgroup$
Define $M_n=frac{1}{n}sum_{i-1}^nX_i$.
$E[M_n]=mu$, $Var[M_n]=frac{sigma_X^2}{n}$
For $mu<0$:
$$P[M_n<frac{mu}{2}]$$
$$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)
$$P[M_n-mu<frac{mu}{2}-mu]$$
$$P[M_n-mu< -frac{mu}{2}]$$
(using Chebyshev inequality)
$$P[|M_n-mu| < -frac{mu}{2}]$$
$$P[|M_n-mu| < -frac{mu}{2}] geq 1 - frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity, so the Probability is $1-0=1$.
For $mu > 0$:
$$P[M_n<frac{mu}{2}]$$
$$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)
$$P[M_n-mu<frac{mu}{2}-mu]$$
$$P[M_n-mu< -frac{mu}{2}]$$
(flip inequality due to division of both sides by -1)
$$P[-M_n+mu > frac{mu}{2}]$$
(using Chebyshev inequality and fact of $|A|=|-A|$)
$$P[|M_n-mu| > frac{mu}{2}]$$
$$P[|M_n-mu| < frac{mu}{2}] leq frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity. So, the probability is 0.
edited Dec 23 '18 at 14:05
answered Dec 23 '18 at 3:50
AvedisAvedis
667
667
$begingroup$
In both cases you should have $(mu/2)^2=mu^2/4$ rather than $mu^2/2$. And it seems our answers agree.
$endgroup$
– angryavian
Dec 23 '18 at 4:20
$begingroup$
You are right and I will fix. I misread your answer. I thought the last statement was a summary of the entire problem, not the last part exclusively. Thank you!
$endgroup$
– Avedis
Dec 23 '18 at 14:04
add a comment |
$begingroup$
In both cases you should have $(mu/2)^2=mu^2/4$ rather than $mu^2/2$. And it seems our answers agree.
$endgroup$
– angryavian
Dec 23 '18 at 4:20
$begingroup$
You are right and I will fix. I misread your answer. I thought the last statement was a summary of the entire problem, not the last part exclusively. Thank you!
$endgroup$
– Avedis
Dec 23 '18 at 14:04
$begingroup$
In both cases you should have $(mu/2)^2=mu^2/4$ rather than $mu^2/2$. And it seems our answers agree.
$endgroup$
– angryavian
Dec 23 '18 at 4:20
$begingroup$
In both cases you should have $(mu/2)^2=mu^2/4$ rather than $mu^2/2$. And it seems our answers agree.
$endgroup$
– angryavian
Dec 23 '18 at 4:20
$begingroup$
You are right and I will fix. I misread your answer. I thought the last statement was a summary of the entire problem, not the last part exclusively. Thank you!
$endgroup$
– Avedis
Dec 23 '18 at 14:04
$begingroup$
You are right and I will fix. I misread your answer. I thought the last statement was a summary of the entire problem, not the last part exclusively. Thank you!
$endgroup$
– Avedis
Dec 23 '18 at 14:04
add a comment |
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