How to derive this simple equality? Let us define L i </msub> &#x225C;<!-- ≜

Aganippe76

Aganippe76

Answered question

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

Beginner2022-07-05Added 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 } .

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