How to derive this simple equality?Let us define L i ≜ log ⁡ ( P...

Aganippe76

Aganippe76

Answered

2022-07-04

How to derive this simple equality?
Let us define
L i log ( P r o b ( x i = + 1 ) P r o b ( x i = 1 ) )
E { x i } P r o b ( x i = + 1 ) P r o b ( x i = 1 )
I need to show that
E { x i } = tanh ( L i / 2 )
I can write
(1a) E { x i } = e l o g ( P r o b ( x i = + 1 ) P r o b ( x i = 1 ) ) (1b) = e L i
if I use the following:
tanh ( z ) = sinh ( z ) cosh ( z ) = e z e z e z + e z = e 2 z 1 e 2 z + 1
and put z = L i / 2, then
(2) tanh ( L i / 2 ) = e L i 1 e l i + 1
how do I arrive at (2) using (1a)

Answer & Explanation

jugf5

jugf5

Expert

2022-07-05Added 18 answers

Where you went wrong is here:
log ( A / B ) log ( A B ) .
Observe that
L i log ( P r o b ( x i = + 1 ) P r o b ( x i = 1 ) )
E { x i } P r o b ( x i = + 1 ) P r o b ( x i = 1 )
tanh ( L i 2 ) = e L i 1 e L i + 1 = P r o b ( x i = + 1 ) P r o b ( x i = 1 ) 1 P r o b ( x i = + 1 ) P r o b ( x i = 1 ) + 1 = P r o b ( x i = + 1 ) P r o b ( x i = 1 ) P r o b ( x i = + 1 ) + P r o b ( x i = 1 ) .
Assuming the only two events are x i = + 1 and x i = 1, we have
P r o b ( x i = + 1 ) + P r o b ( x i = 1 ) = 1.
Therefore,
tanh ( L i 2 ) = E { x i } .

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?