p(s=1|x) = 1/(1+exp(-x/T))) ,where ‘s’ is the output given the ac – McqOptions

MCQOPTIONS

1.	p(s=1\|x) = 1/(1+exp(-x/T))) ,where ‘s’ is the output given the activation ‘x’ is a?
A.	hopfield network
B.	sigma network
C.	stochastic network
D.	none of the mentioned
Answer» D. none of the mentioned

Discussion

No Comment Found

Related MCQs

Where does a stochastic network exhibits stable states ?
When activation value is determined by using the average of fluctuations of outputs from other units, it is known as?
Is it possible in stochastic network that average state of network doesn’t change with time?
In case of stochastic update, what kind of equilibrium is reached?
p(s=1|x) = 1/(1+exp(-x/T))) ,where ‘s’ is the output given the activation ‘x’ is a?
WHEN_ACTIVATION_VALUE_IS_DETERMINED_BY_USING_THE_AVERAGE_OF_FLUCTUATIONS_OF_OUTPUTS_FROM_OTHER_UNITS,_IT_IS_KNOWN_AS??$
Where_does_a_stochastic_network_exhibits_stable_states_?$
Can networks with symmetric weight reach thermal equilibrium?
What can be the possible reason for thermal equilibrium in stochastic networks?
Is it possible in stochastic network that average state of network doesn‚Äö√Ñ√∂‚àö√ë‚àö¬•t change with time?$