Understanding why variance of the standard normal distribution equals one intuitively
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Can anyone explain to me why the variance of the standard normal distribution is 1? I am trying to understand the mechanism behind standardising random variable. While I know minus the variable by the mean is like shifting the graph to make it centre at the origin, I don't know why dividing it by SD makes the variable having SD = 1 as well
calculus linear-algebra statistics linear-transformations normal-distribution
asked Jan 12 at 11:04
Stephen FongStephen Fong
114
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add a comment |
$begingroup$
Can anyone explain to me why the variance of the standard normal distribution is 1? I am trying to understand the mechanism behind standardising random variable. While I know minus the variable by the mean is like shifting the graph to make it centre at the origin, I don't know why dividing it by SD makes the variable having SD = 1 as well
calculus linear-algebra statistics linear-transformations normal-distribution
asked Jan 12 at 11:04
Stephen FongStephen Fong
114
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4
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It's the definition...if you have a normal distribution with any non-zero standard deviation you can rescale to get $sigma =1 $ so it's really just a matter of units.
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– lulu
Jan 12 at 11:05
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If you view the SD as a thing that tells you how dispersed your distribution is around the mean, then you can understand why dividing your variable by a constant will divide your SD by this constant. So now just divide by the SD itself. Your SD has now become 1.
$endgroup$
– Bermudes
Jan 12 at 11:17
add a comment |
$begingroup$
Can anyone explain to me why the variance of the standard normal distribution is 1? I am trying to understand the mechanism behind standardising random variable. While I know minus the variable by the mean is like shifting the graph to make it centre at the origin, I don't know why dividing it by SD makes the variable having SD = 1 as well
calculus linear-algebra statistics linear-transformations normal-distribution
asked Jan 12 at 11:04
Stephen FongStephen Fong
114
$endgroup$
Can anyone explain to me why the variance of the standard normal distribution is 1? I am trying to understand the mechanism behind standardising random variable. While I know minus the variable by the mean is like shifting the graph to make it centre at the origin, I don't know why dividing it by SD makes the variable having SD = 1 as well
calculus linear-algebra statistics linear-transformations normal-distribution
calculus linear-algebra statistics linear-transformations normal-distribution
asked Jan 12 at 11:04
Stephen FongStephen Fong
114
asked Jan 12 at 11:04
Stephen FongStephen Fong
114
asked Jan 12 at 11:04
Stephen FongStephen Fong
114
asked Jan 12 at 11:04
Stephen FongStephen Fong
114
asked Jan 12 at 11:04
Stephen FongStephen Fong
114
114
4
$begingroup$
It's the definition...if you have a normal distribution with any non-zero standard deviation you can rescale to get $sigma =1 $ so it's really just a matter of units.
$endgroup$
– lulu
Jan 12 at 11:05
$begingroup$
If you view the SD as a thing that tells you how dispersed your distribution is around the mean, then you can understand why dividing your variable by a constant will divide your SD by this constant. So now just divide by the SD itself. Your SD has now become 1.
$endgroup$
– Bermudes
Jan 12 at 11:17
add a comment |
4
$begingroup$
It's the definition...if you have a normal distribution with any non-zero standard deviation you can rescale to get $sigma =1 $ so it's really just a matter of units.
$endgroup$
– lulu
Jan 12 at 11:05
$begingroup$
If you view the SD as a thing that tells you how dispersed your distribution is around the mean, then you can understand why dividing your variable by a constant will divide your SD by this constant. So now just divide by the SD itself. Your SD has now become 1.
$endgroup$
– Bermudes
Jan 12 at 11:17
4
4
$begingroup$
It's the definition...if you have a normal distribution with any non-zero standard deviation you can rescale to get $sigma =1 $ so it's really just a matter of units.
$endgroup$
– lulu
Jan 12 at 11:05
$begingroup$
It's the definition...if you have a normal distribution with any non-zero standard deviation you can rescale to get $sigma =1 $ so it's really just a matter of units.
$endgroup$
– lulu
Jan 12 at 11:05
$begingroup$
If you view the SD as a thing that tells you how dispersed your distribution is around the mean, then you can understand why dividing your variable by a constant will divide your SD by this constant. So now just divide by the SD itself. Your SD has now become 1.
$endgroup$
– Bermudes
Jan 12 at 11:17
$begingroup$
If you view the SD as a thing that tells you how dispersed your distribution is around the mean, then you can understand why dividing your variable by a constant will divide your SD by this constant. So now just divide by the SD itself. Your SD has now become 1.
$endgroup$
– Bermudes
Jan 12 at 11:17
add a comment |
2 Answers
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The variance of standard normal distribution is $1$ by definition.
Concerning standardizing: if $X$ has a distribution with standard deviation $sigma_Xneq0$ or equivalently with variance $sigma_X^2$ then for every constant $c$ (also $c=mathbb EX$) we have $mathsf{Var}left(frac{X-c}{sigma_X}right)=1$ according to the rule:$$mathsf{Var}(aY+b)=a^2mathsf{Var}Y$$
Can you deduce this rule yourself?
Applying it on $Y=frac{X-c}{sigma_X}$ we get: $$mathsf{Var}(sigma_X^{-1}X+(-sigma_X^{-1}c))=sigma_X^{-2}mathsf{Var}X=sigma_X^{-2}sigma_X^{2}=1$$
This means that we can write $X=sigma_XU+mu_X$ where $U:=frac{X-mu_X}{sigma_X}$ has mean $0$ and variance $1$.
answered Jan 12 at 11:21
drhabdrhab
99.9k544130
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Let $Xsim N(mu,sigma^2)$ and $Z=frac{X-mu}{sigma}$, then $Zsim N(0,1)$, because:
$$mathbb Eleft(frac{X-mu}{sigma}right)=frac{1}{sigma}cdot mathbb E(X-mu)=frac1{sigma}cdot mathbb E(X)-frac{mu}{sigma}=0;\
sigma^2left(frac{X-mu}{sigma}right)=frac{1}{sigma^2}cdot sigma^2(X-mu)=frac1{sigma^2}cdot sigma^2(X)=1.$$
answered Jan 12 at 12:11
farruhotafarruhota
19.9k2738
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$begingroup$
The variance of standard normal distribution is $1$ by definition.
Concerning standardizing: if $X$ has a distribution with standard deviation $sigma_Xneq0$ or equivalently with variance $sigma_X^2$ then for every constant $c$ (also $c=mathbb EX$) we have $mathsf{Var}left(frac{X-c}{sigma_X}right)=1$ according to the rule:$$mathsf{Var}(aY+b)=a^2mathsf{Var}Y$$
Can you deduce this rule yourself?
Applying it on $Y=frac{X-c}{sigma_X}$ we get: $$mathsf{Var}(sigma_X^{-1}X+(-sigma_X^{-1}c))=sigma_X^{-2}mathsf{Var}X=sigma_X^{-2}sigma_X^{2}=1$$
This means that we can write $X=sigma_XU+mu_X$ where $U:=frac{X-mu_X}{sigma_X}$ has mean $0$ and variance $1$.
answered Jan 12 at 11:21
drhabdrhab
99.9k544130
$endgroup$
add a comment |
$begingroup$
The variance of standard normal distribution is $1$ by definition.
Concerning standardizing: if $X$ has a distribution with standard deviation $sigma_Xneq0$ or equivalently with variance $sigma_X^2$ then for every constant $c$ (also $c=mathbb EX$) we have $mathsf{Var}left(frac{X-c}{sigma_X}right)=1$ according to the rule:$$mathsf{Var}(aY+b)=a^2mathsf{Var}Y$$
Can you deduce this rule yourself?
Applying it on $Y=frac{X-c}{sigma_X}$ we get: $$mathsf{Var}(sigma_X^{-1}X+(-sigma_X^{-1}c))=sigma_X^{-2}mathsf{Var}X=sigma_X^{-2}sigma_X^{2}=1$$
This means that we can write $X=sigma_XU+mu_X$ where $U:=frac{X-mu_X}{sigma_X}$ has mean $0$ and variance $1$.
answered Jan 12 at 11:21
drhabdrhab
99.9k544130
$endgroup$
add a comment |
$begingroup$
The variance of standard normal distribution is $1$ by definition.
Concerning standardizing: if $X$ has a distribution with standard deviation $sigma_Xneq0$ or equivalently with variance $sigma_X^2$ then for every constant $c$ (also $c=mathbb EX$) we have $mathsf{Var}left(frac{X-c}{sigma_X}right)=1$ according to the rule:$$mathsf{Var}(aY+b)=a^2mathsf{Var}Y$$
Can you deduce this rule yourself?
Applying it on $Y=frac{X-c}{sigma_X}$ we get: $$mathsf{Var}(sigma_X^{-1}X+(-sigma_X^{-1}c))=sigma_X^{-2}mathsf{Var}X=sigma_X^{-2}sigma_X^{2}=1$$
This means that we can write $X=sigma_XU+mu_X$ where $U:=frac{X-mu_X}{sigma_X}$ has mean $0$ and variance $1$.
answered Jan 12 at 11:21
drhabdrhab
99.9k544130
$endgroup$
The variance of standard normal distribution is $1$ by definition.
Concerning standardizing: if $X$ has a distribution with standard deviation $sigma_Xneq0$ or equivalently with variance $sigma_X^2$ then for every constant $c$ (also $c=mathbb EX$) we have $mathsf{Var}left(frac{X-c}{sigma_X}right)=1$ according to the rule:$$mathsf{Var}(aY+b)=a^2mathsf{Var}Y$$
Can you deduce this rule yourself?
Applying it on $Y=frac{X-c}{sigma_X}$ we get: $$mathsf{Var}(sigma_X^{-1}X+(-sigma_X^{-1}c))=sigma_X^{-2}mathsf{Var}X=sigma_X^{-2}sigma_X^{2}=1$$
This means that we can write $X=sigma_XU+mu_X$ where $U:=frac{X-mu_X}{sigma_X}$ has mean $0$ and variance $1$.
answered Jan 12 at 11:21
drhabdrhab
99.9k544130
answered Jan 12 at 11:21
drhabdrhab
99.9k544130
answered Jan 12 at 11:21
drhabdrhab
99.9k544130
answered Jan 12 at 11:21
drhabdrhab
99.9k544130
99.9k544130
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$begingroup$
Let $Xsim N(mu,sigma^2)$ and $Z=frac{X-mu}{sigma}$, then $Zsim N(0,1)$, because:
$$mathbb Eleft(frac{X-mu}{sigma}right)=frac{1}{sigma}cdot mathbb E(X-mu)=frac1{sigma}cdot mathbb E(X)-frac{mu}{sigma}=0;\
sigma^2left(frac{X-mu}{sigma}right)=frac{1}{sigma^2}cdot sigma^2(X-mu)=frac1{sigma^2}cdot sigma^2(X)=1.$$
answered Jan 12 at 12:11
farruhotafarruhota
19.9k2738
$endgroup$
add a comment |
$begingroup$
Let $Xsim N(mu,sigma^2)$ and $Z=frac{X-mu}{sigma}$, then $Zsim N(0,1)$, because:
$$mathbb Eleft(frac{X-mu}{sigma}right)=frac{1}{sigma}cdot mathbb E(X-mu)=frac1{sigma}cdot mathbb E(X)-frac{mu}{sigma}=0;\
sigma^2left(frac{X-mu}{sigma}right)=frac{1}{sigma^2}cdot sigma^2(X-mu)=frac1{sigma^2}cdot sigma^2(X)=1.$$
answered Jan 12 at 12:11
farruhotafarruhota
19.9k2738
$endgroup$
add a comment |
$begingroup$
Let $Xsim N(mu,sigma^2)$ and $Z=frac{X-mu}{sigma}$, then $Zsim N(0,1)$, because:
$$mathbb Eleft(frac{X-mu}{sigma}right)=frac{1}{sigma}cdot mathbb E(X-mu)=frac1{sigma}cdot mathbb E(X)-frac{mu}{sigma}=0;\
sigma^2left(frac{X-mu}{sigma}right)=frac{1}{sigma^2}cdot sigma^2(X-mu)=frac1{sigma^2}cdot sigma^2(X)=1.$$
answered Jan 12 at 12:11
farruhotafarruhota
19.9k2738
$endgroup$
Let $Xsim N(mu,sigma^2)$ and $Z=frac{X-mu}{sigma}$, then $Zsim N(0,1)$, because:
$$mathbb Eleft(frac{X-mu}{sigma}right)=frac{1}{sigma}cdot mathbb E(X-mu)=frac1{sigma}cdot mathbb E(X)-frac{mu}{sigma}=0;\
sigma^2left(frac{X-mu}{sigma}right)=frac{1}{sigma^2}cdot sigma^2(X-mu)=frac1{sigma^2}cdot sigma^2(X)=1.$$
answered Jan 12 at 12:11
farruhotafarruhota
19.9k2738
answered Jan 12 at 12:11
farruhotafarruhota
19.9k2738
answered Jan 12 at 12:11
farruhotafarruhota
19.9k2738
answered Jan 12 at 12:11
farruhotafarruhota
19.9k2738
19.9k2738
add a comment |
add a comment |
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$begingroup$
It's the definition...if you have a normal distribution with any non-zero standard deviation you can rescale to get $sigma =1 $ so it's really just a matter of units.
$endgroup$
– lulu
Jan 12 at 11:05
$begingroup$
If you view the SD as a thing that tells you how dispersed your distribution is around the mean, then you can understand why dividing your variable by a constant will divide your SD by this constant. So now just divide by the SD itself. Your SD has now become 1.
$endgroup$
– Bermudes
Jan 12 at 11:17