Wrong proof?

[nosit]

I am watching the derivation of the Fisher Formula from a nice channel on Youtube;
until I see this. Is this wrong or am I missing something?

As (1+pi) has been multiplied in both sides, in the RHS of the formula we would has to be left 1+i-1+1

 

[pka]

Their, the printed, are correct.
\(\large r(1+\pi)=1+i-(1+\pi)\)
\(\large r+r\pi=1+i-1{\Large\bf{-}}\pi\)

 

[nosit]

@pka thank you.

But what about the denominator (1+pi) which was already in the RHS?
We would have in the RHS (1+pi)/(1+pi) which is 1
So the RHS would be 1+i-1 from my point o view.

 

[Dr.Peterson]

I think you are forgetting to multiply the entire RHS, and just multiplying the first term. Here are the details:

[MATH]r=\frac{1+i}{1+\pi}-1[/MATH]​

​

[MATH]r(1+\pi)=\left[\frac{1+i}{1+\pi}-1\right](1+\pi)[/MATH]​

​

[MATH]r(1+\pi)=\left[\frac{1+i}{1+\pi}\right](1+\pi)-1(1+\pi)[/MATH]​

​

[MATH]r(1+\pi)=\frac{(1+i)(1+\pi)}{1+\pi}-(1+\pi)[/MATH]​

​

[MATH]r+r\pi=(1+i)-(1+\pi)[/MATH]​

​

[MATH]r+r\pi=1+i-1-\pi[/MATH]​

​

[MATH]r+r\pi=i-\pi[/MATH]​

 

[nosit]

Thank you @Dr.Peterson

 

[lex]

When you multiply the RHS by [MATH](1+\pi)[/MATH] I wonder if you were thinking that [MATH]\frac{1+\pi}{1+\pi}[/MATH] gives you 1 which is ADDED to the end giving 1+i - 1 +1
But the 1 is actually multiplying the (1+i)

[MATH]r= \boxed{(1+i)\times\frac{1}{1+\pi}} - \boxed{1}\\ r(1+\pi)=\boxed{(1+i)\times\frac{1}{1+\pi}\times(1+\pi)} - \boxed{1\times(1+\pi)}\\ r(1+\pi)=\boxed{(1+i)\times\frac{1+\pi}{1+\pi}} - \boxed{(1+\pi)}\\ r(1+\pi)=\boxed{(1+i)\times 1} - \boxed{(1+\pi)}\\ r(1+\pi)=(1+i) - (1+\pi)\\ r(1+\pi)=1+i - 1-\pi\\ r(1+\pi)=i-\pi[/MATH]...

 

[nosit]

Thank you @lex
Taking advantage of your reply

I was wondering if mathematically a distributive to the other side is allowed (where I highlighted in yellow). Is it?
It does not seem to be, because the result would be different, it would be:

-1-i+1+pi
pi-i

which is different...

 

[lex]

I was wondering if mathematically a distributive to the other side is allowed (where I highlighted in yellow). Is it?
It does not seem to be, because the result would be different, it would be:
-1-i+1+pi
pi-i
which is different...

I think you mean, can they commute? The answer is no, because subtraction is not commutative:
[MATH]a-b\not\equiv b-a[/MATH]Addition is commutative, so you could think of it as:
[MATH]a+ \text{-}b \equiv \text{-}b+a[/MATH]i.e. [MATH](1+i)+\text{-}(1+\pi)=\text{-}(1+\pi)+(1+i)[/MATH]

 

[nosit]

Thank you @lex !

 

[JeffM]

Two things

The demonstration is unnecessarily obscure.

[MATH]r = \dfrac{\dfrac{1 + i}{1 + \pi} - 1}{1} = \dfrac{1 + i}{1 + \pi } - 1 = \dfrac{1 + i - (1 + \pi )}{1 + \pi} \implies [/MATH]
[MATH]r = \dfrac{i - \pi }{1 + \pi}.[/MATH]
Now it is easy to see

[MATH]r(1 + \pi)= \dfrac{i - \pi}{1 + \pi} * (1 + \pi) \implies r + r \pi = i - \pi \implies i = r + \pi + r \pi.[/MATH]
Second, if r and pi are small, r times pi is negligible.

[MATH]i = 0.1 \text { and } \pi = 0.05 \implies r = \dfrac{0.1 - 0.5}{1.05} \approx 0.047619.[/MATH]
[MATH]0.05 * 0.047619 = 0.00238095, \text { just over 2 basis points.} [/MATH]
Unless we are dealing with very rapid inflation, the error in the text book version is not worth bothering about because it is swamped by the uncertainty in the expected inflation rate.[/MATH]