Given a sequence ${ X_{n} : n in mathbb{N} }$ of standard normally distributed random variables, what is the...
up vote
0
down vote
favorite
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 at 20:25
S. Crim
9210
add a comment |
up vote
0
down vote
favorite
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 at 20:25
S. Crim
9210
If they are independent, then the sum is a normal random variable.
– gd1035
Nov 20 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 at 20:30
Hint: The variance of a sum of independent random variables is the sum of the variances.
– Robert Israel
Nov 20 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 at 20:43
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 at 20:25
S. Crim
9210
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 at 20:25
S. Crim
9210
asked Nov 20 at 20:25
S. Crim
9210
asked Nov 20 at 20:25
S. Crim
9210
asked Nov 20 at 20:25
S. Crim
9210
asked Nov 20 at 20:25
S. Crim
9210
9210
If they are independent, then the sum is a normal random variable.
– gd1035
Nov 20 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 at 20:30
Hint: The variance of a sum of independent random variables is the sum of the variances.
– Robert Israel
Nov 20 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 at 20:43
add a comment |
If they are independent, then the sum is a normal random variable.
– gd1035
Nov 20 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 at 20:30
Hint: The variance of a sum of independent random variables is the sum of the variances.
– Robert Israel
Nov 20 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 at 20:43
If they are independent, then the sum is a normal random variable.
– gd1035
Nov 20 at 20:28
If they are independent, then the sum is a normal random variable.
– gd1035
Nov 20 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 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 at 20:30
Hint: The variance of a sum of independent random variables is the sum of the variances.
– Robert Israel
Nov 20 at 20:34
Hint: The variance of a sum of independent random variables is the sum of the variances.
– Robert Israel
Nov 20 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 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 at 20:43
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
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 at 21:00
Aditya Dua
5006
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
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 at 21:00
Aditya Dua
5006
add a comment |
up vote
2
down vote
accepted
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 at 21:00
Aditya Dua
5006
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
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 at 21:00
Aditya Dua
5006
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 at 21:00
Aditya Dua
5006
answered Nov 20 at 21:00
Aditya Dua
5006
answered Nov 20 at 21:00
Aditya Dua
5006
answered Nov 20 at 21:00
Aditya Dua
5006
5006
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006850%2fgiven-a-sequence-x-n-n-in-mathbbn-of-standard-normally-distribut%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
If they are independent, then the sum is a normal random variable.
– gd1035
Nov 20 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 at 20:30
Hint: The variance of a sum of independent random variables is the sum of the variances.
– Robert Israel
Nov 20 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 at 20:43