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Glen_b
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That this is not the case can easily be seen simply by computing the cdf for a different distribution than the normal.

So for example, consider a standard normal (i.e. let's fix $\mu$ and $\sigma$ at 0 and 1, without loss of generality$^\dagger$).

Let's try a uniform on $(-\sqrt{3},\sqrt{3})$ (which also has mean 0 and sd 1) for comparison:

![enter image description here

We can see that it's not the case that $\Phi\leq F$ everywhere. There are places where either will exceed the other.

$\dagger$ we can do the same for the general $\mu,\sigma$ case by linearly scaling this example.

Indeed, if you're going to have the means be equal and you're going to make $F\gt\Phi$ in some region, it's going to have to be greatersmaller somewhere else.

That this is not the case can easily be seen simply by computing the cdf for a different distribution than the normal.

So for example, consider a standard normal (i.e. let's fix $\mu$ and $\sigma$ at 0 and 1, without loss of generality$^\dagger$).

Let's try a uniform on $(-\sqrt{3},\sqrt{3})$ (which also has mean 0 and sd 1) for comparison:

![enter image description here

We can see that it's not the case that $\Phi\leq F$ everywhere. There are places where either will exceed the other.

$\dagger$ we can do the same for the general $\mu,\sigma$ case by linearly scaling this example.

Indeed, if you're going to have the means be equal and you're going to make $F\gt\Phi$ in some region, it's going to have to be greater somewhere else.

That this is not the case can easily be seen simply by computing the cdf for a different distribution than the normal.

So for example, consider a standard normal (i.e. let's fix $\mu$ and $\sigma$ at 0 and 1, without loss of generality$^\dagger$).

Let's try a uniform on $(-\sqrt{3},\sqrt{3})$ (which also has mean 0 and sd 1) for comparison:

![enter image description here

We can see that it's not the case that $\Phi\leq F$ everywhere. There are places where either will exceed the other.

$\dagger$ we can do the same for the general $\mu,\sigma$ case by linearly scaling this example.

Indeed, if you're going to have the means be equal and you're going to make $F\gt\Phi$ in some region, it's going to have to be smaller somewhere else.

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Glen_b
  • 290.5k
  • 37
  • 652
  • 1.1k

That this is not the case can easily be seen simply by computing the cdf for a different distribution than the normal.

So for example, consider a standard normal (i.e. let's fix $\mu$ and $\sigma$ at 0 and 1, without loss of generality$^\dagger$).

Let's try a uniform on $(-\sqrt{3},\sqrt{3})$ (which also has mean 0 and sd 1) for comparison:

![enter image description here

We can see that it's not the case that $\Phi\leq F$ everywhere. There are places where either will exceed the other.

$\dagger$ we can do the same for the general $\mu,\sigma$ case by linearly scaling this example.

Indeed, if you're going to have the means be equal and you're going to make $F\gt\Phi$ in some region, it's going to have to be greater somewhere else.