@goldpiggy Many thanks! Finally understood it!
?
Hello. I am still not able to understand clearly how these 2 equations are related. Can you please explain, how for a particular observation i, the probability y given x is related to the entropy definition overall classes?
Thank you for your response. My question was more specifically why
is same as
l(y,y_hat)
Is this because y when 1-hot encoded has only single position with 1 and hence when we sum up the y * log(y_hat) over the entire class, we are left with the probability y_hat corresponding to true y. Please advise.
@Abinash_Sahu
l (y，y _ hat)
Cross entropy loss
Only one type of these losses we often use.
https://ml-cheatsheet.readthedocs.io/en/latest/loss_functions.html
Q1.2. Compute the variance of the distribution given by softmax(𝐨)softmax(o) and show that it matches the second derivative computed above.
Can someone point me in the right direction? I tried to use Var[𝑋]=𝐸[(𝑋−𝐸[𝑋])^2]=𝐸[𝑋^2]−𝐸[𝑋]^2 to find the variance but I ended up having the term 1/q^2… it doesn’t look like the second derivative from Q1.1.
Thanks!
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It appears that there is a reference that remained unresolved:
:eqref: eq_l_cross_entropy
in 3.4.5.3
to keep unified form,should the yj in later two equations should have an upper right mark (i) ?
I cannot understand the equation in 3.4.9
I may have an explanation considering equivalence between log likehood and cross-entropy.
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Looks like you have der(E[X]^2), but what about E[X^2]? Recall:
Var[X] = E[X^2] - E[X]^2
3.1 is strait forward to show.
I’m having trouble with 3.2 and 3.3:
3.2:
Show:
⎛ a b⎞
log⎝𝜆⋅ℯ + 𝜆⋅ℯ ⎠
──────────────── > Max(a, b)
𝜆
Assume:
a > b
𝜆 > 0
(Max(a,b) -> a, b/c a > b)
⎛ a b⎞
log⎝𝜆⋅ℯ + 𝜆⋅ℯ ⎠
──────────────── > a
𝜆
⎛ a b⎞
log⎝𝜆⋅ℯ + 𝜆⋅ℯ ⎠ > 𝜆a
(exp both sides)
a b 𝜆a
𝜆⋅ℯ + 𝜆⋅ℯ > ℯ
LHS !> RHS
and 3.3:
I did the calculus and the limit looked like it was going to zero (instead of max(a,b)) so I coded up the function in numpy to check, and indeed it appears to go to 0 instead of 4 in this case (a=2, b=4).
[nav] In [478]: real_softmax = lambda x: 1/x * np.log(x*np.exp(2) + x*np.exp(4))
[ins] In [479]: real_softmax(.1)
Out[479]: 18.24342918048927
[nav] In [480]: real_softmax(1)
Out[480]: 4.126928011042972
[ins] In [481]: real_softmax(10)
Out[481]: 0.6429513104037019
[ins] In [482]: real_softmax(1000)
Out[482]: 0.01103468329002511
[ins] In [483]: real_softmax(100000)
Out[483]: 0.000156398534760132
Please advise
thanks.but i use the definition of variance to derive while your advice is to use inference of variance to do that.both are same in fact
i think you have a mistake at the usage of ∑a/b != ∑a/∑b but ∑a / b as the denominator is public
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