Mixing
Given a sequence ${ X_{n} : n in mathbb{N} }$ of standard normally distributed random variables, what is the...
I'm looking for a general formula that can be used to compute the probability that a sum of standard normal random variables is above a certain constant $a$. So for example the probability that $X_{1}+X_{2}+X_{3}+X_{4}<3$. I've tried to play around with the distribution of a sum of standard normal distributed random variables but wasn't able to come up with a real formula. Any help is appreciated.
probability probability-theory statistics probability-distributions normal-distribution
asked Nov 20 '18 at 20:25
S. Crim
11710
add a comment |
I'm looking for a general formula that can be used to compute the probability that a sum of standard normal random variables is above a certain constant $a$. So for example the probability that $X_{1}+X_{2}+X_{3}+X_{4}<3$. I've tried to play around with the distribution of a sum of standard normal distributed random variables but wasn't able to come up with a real formula. Any help is appreciated.
probability probability-theory statistics probability-distributions normal-distribution
asked Nov 20 '18 at 20:25
S. Crim
11710
If they are independent, then the sum is a normal random variable.
– gd1035
Nov 20 '18 at 20:28
@gd1035 They are indeed considered independent. In that case, what is the probability of the sum being smaller than some constant $a$?
– S. Crim
Nov 20 '18 at 20:30
Hint: The variance of a sum of independent random variables is the sum of the variances.
– Robert Israel
Nov 20 '18 at 20:34
@RobertIsrael Yeah I know that the $sum^n_{i=1} X_{i} sim N(0,n)$ but I remain unable to turn this into something that can tell me $mathbb{P}(sum^n_{i=1} X_{i}>a)$. Any hints for that?
– S. Crim
Nov 20 '18 at 20:43
add a comment |
I'm looking for a general formula that can be used to compute the probability that a sum of standard normal random variables is above a certain constant $a$. So for example the probability that $X_{1}+X_{2}+X_{3}+X_{4}<3$. I've tried to play around with the distribution of a sum of standard normal distributed random variables but wasn't able to come up with a real formula. Any help is appreciated.
probability probability-theory statistics probability-distributions normal-distribution
asked Nov 20 '18 at 20:25
S. Crim
11710
I'm looking for a general formula that can be used to compute the probability that a sum of standard normal random variables is above a certain constant $a$. So for example the probability that $X_{1}+X_{2}+X_{3}+X_{4}<3$. I've tried to play around with the distribution of a sum of standard normal distributed random variables but wasn't able to come up with a real formula. Any help is appreciated.
probability probability-theory statistics probability-distributions normal-distribution
probability probability-theory statistics probability-distributions normal-distribution
asked Nov 20 '18 at 20:25
S. Crim
11710
asked Nov 20 '18 at 20:25
S. Crim
11710
asked Nov 20 '18 at 20:25
S. Crim
11710
asked Nov 20 '18 at 20:25
S. Crim
11710
asked Nov 20 '18 at 20:25
S. Crim
11710
11710
If they are independent, then the sum is a normal random variable.
– gd1035
Nov 20 '18 at 20:28
@gd1035 They are indeed considered independent. In that case, what is the probability of the sum being smaller than some constant $a$?
– S. Crim
Nov 20 '18 at 20:30
Hint: The variance of a sum of independent random variables is the sum of the variances.
– Robert Israel
Nov 20 '18 at 20:34
@RobertIsrael Yeah I know that the $sum^n_{i=1} X_{i} sim N(0,n)$ but I remain unable to turn this into something that can tell me $mathbb{P}(sum^n_{i=1} X_{i}>a)$. Any hints for that?
– S. Crim
Nov 20 '18 at 20:43
add a comment |
If they are independent, then the sum is a normal random variable.
– gd1035
Nov 20 '18 at 20:28
@gd1035 They are indeed considered independent. In that case, what is the probability of the sum being smaller than some constant $a$?
– S. Crim
Nov 20 '18 at 20:30
Hint: The variance of a sum of independent random variables is the sum of the variances.
– Robert Israel
Nov 20 '18 at 20:34
@RobertIsrael Yeah I know that the $sum^n_{i=1} X_{i} sim N(0,n)$ but I remain unable to turn this into something that can tell me $mathbb{P}(sum^n_{i=1} X_{i}>a)$. Any hints for that?
– S. Crim
Nov 20 '18 at 20:43
If they are independent, then the sum is a normal random variable.
– gd1035
Nov 20 '18 at 20:28
If they are independent, then the sum is a normal random variable.
– gd1035
Nov 20 '18 at 20:28
@gd1035 They are indeed considered independent. In that case, what is the probability of the sum being smaller than some constant $a$?
– S. Crim
Nov 20 '18 at 20:30
@gd1035 They are indeed considered independent. In that case, what is the probability of the sum being smaller than some constant $a$?
– S. Crim
Nov 20 '18 at 20:30
Hint: The variance of a sum of independent random variables is the sum of the variances.
– Robert Israel
Nov 20 '18 at 20:34
Hint: The variance of a sum of independent random variables is the sum of the variances.
– Robert Israel
Nov 20 '18 at 20:34
@RobertIsrael Yeah I know that the $sum^n_{i=1} X_{i} sim N(0,n)$ but I remain unable to turn this into something that can tell me $mathbb{P}(sum^n_{i=1} X_{i}>a)$. Any hints for that?
– S. Crim
Nov 20 '18 at 20:43
@RobertIsrael Yeah I know that the $sum^n_{i=1} X_{i} sim N(0,n)$ but I remain unable to turn this into something that can tell me $mathbb{P}(sum^n_{i=1} X_{i}>a)$. Any hints for that?
– S. Crim
Nov 20 '18 at 20:43
add a comment |
1 Answer
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For a R.V. $Y sim N(0,sigma^2)$, by definition:
$P(Y leq y) = int_{-infty}^y frac{1}{sigma sqrt{2 pi}}e^{-x^2/2 sigma^2}dx$.
Let's substitute $u = x/sigma$ to get:
$P(Y leq y) = int_{-infty}^{y/sigma} frac{1}{sqrt{2 pi}} e^{-u^2/2} du = Phi left(frac{y}{sigma} right)$
where $Phi$ is the cumulative distribution function of a standard normal variable (which has zero mean and unit variance). You can evaluate $Phi$ in any mathematical package or from printed or online Z tables.
In your specific case, as many have pointed out, $ Y = sum_{i=1}^n X_i$ has mean 0 and variance $sigma^2 = n$, which means you need to evaluate $Phi(y/n)$ to compute $P(Y leq y)$.
answered Nov 20 '18 at 21:00
Aditya Dua
83918
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
For a R.V. $Y sim N(0,sigma^2)$, by definition:
$P(Y leq y) = int_{-infty}^y frac{1}{sigma sqrt{2 pi}}e^{-x^2/2 sigma^2}dx$.
Let's substitute $u = x/sigma$ to get:
$P(Y leq y) = int_{-infty}^{y/sigma} frac{1}{sqrt{2 pi}} e^{-u^2/2} du = Phi left(frac{y}{sigma} right)$
where $Phi$ is the cumulative distribution function of a standard normal variable (which has zero mean and unit variance). You can evaluate $Phi$ in any mathematical package or from printed or online Z tables.
In your specific case, as many have pointed out, $ Y = sum_{i=1}^n X_i$ has mean 0 and variance $sigma^2 = n$, which means you need to evaluate $Phi(y/n)$ to compute $P(Y leq y)$.
answered Nov 20 '18 at 21:00
Aditya Dua
83918
add a comment |
For a R.V. $Y sim N(0,sigma^2)$, by definition:
$P(Y leq y) = int_{-infty}^y frac{1}{sigma sqrt{2 pi}}e^{-x^2/2 sigma^2}dx$.
Let's substitute $u = x/sigma$ to get:
$P(Y leq y) = int_{-infty}^{y/sigma} frac{1}{sqrt{2 pi}} e^{-u^2/2} du = Phi left(frac{y}{sigma} right)$
where $Phi$ is the cumulative distribution function of a standard normal variable (which has zero mean and unit variance). You can evaluate $Phi$ in any mathematical package or from printed or online Z tables.
In your specific case, as many have pointed out, $ Y = sum_{i=1}^n X_i$ has mean 0 and variance $sigma^2 = n$, which means you need to evaluate $Phi(y/n)$ to compute $P(Y leq y)$.
answered Nov 20 '18 at 21:00
Aditya Dua
83918
add a comment |
For a R.V. $Y sim N(0,sigma^2)$, by definition:
$P(Y leq y) = int_{-infty}^y frac{1}{sigma sqrt{2 pi}}e^{-x^2/2 sigma^2}dx$.
Let's substitute $u = x/sigma$ to get:
$P(Y leq y) = int_{-infty}^{y/sigma} frac{1}{sqrt{2 pi}} e^{-u^2/2} du = Phi left(frac{y}{sigma} right)$
where $Phi$ is the cumulative distribution function of a standard normal variable (which has zero mean and unit variance). You can evaluate $Phi$ in any mathematical package or from printed or online Z tables.
In your specific case, as many have pointed out, $ Y = sum_{i=1}^n X_i$ has mean 0 and variance $sigma^2 = n$, which means you need to evaluate $Phi(y/n)$ to compute $P(Y leq y)$.
answered Nov 20 '18 at 21:00
Aditya Dua
83918
For a R.V. $Y sim N(0,sigma^2)$, by definition:
$P(Y leq y) = int_{-infty}^y frac{1}{sigma sqrt{2 pi}}e^{-x^2/2 sigma^2}dx$.
Let's substitute $u = x/sigma$ to get:
$P(Y leq y) = int_{-infty}^{y/sigma} frac{1}{sqrt{2 pi}} e^{-u^2/2} du = Phi left(frac{y}{sigma} right)$
where $Phi$ is the cumulative distribution function of a standard normal variable (which has zero mean and unit variance). You can evaluate $Phi$ in any mathematical package or from printed or online Z tables.
In your specific case, as many have pointed out, $ Y = sum_{i=1}^n X_i$ has mean 0 and variance $sigma^2 = n$, which means you need to evaluate $Phi(y/n)$ to compute $P(Y leq y)$.
answered Nov 20 '18 at 21:00
Aditya Dua
83918
answered Nov 20 '18 at 21:00
Aditya Dua
83918
answered Nov 20 '18 at 21:00
Aditya Dua
83918
answered Nov 20 '18 at 21:00
Aditya Dua
83918
83918
add a comment |
add a comment |
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If they are independent, then the sum is a normal random variable.
– gd1035
Nov 20 '18 at 20:28
@gd1035 They are indeed considered independent. In that case, what is the probability of the sum being smaller than some constant $a$?
– S. Crim
Nov 20 '18 at 20:30
Hint: The variance of a sum of independent random variables is the sum of the variances.
– Robert Israel
Nov 20 '18 at 20:34
@RobertIsrael Yeah I know that the $sum^n_{i=1} X_{i} sim N(0,n)$ but I remain unable to turn this into something that can tell me $mathbb{P}(sum^n_{i=1} X_{i}>a)$. Any hints for that?
– S. Crim
Nov 20 '18 at 20:43