Dear friends,
I have a problem with the derivation of the expected value of a stochastic variable. my question is if the following relation is correct or not:
tau (t) is stochastic variable follow Bernoulli distribution if :
\frac{d}{dt} E\{\tau(t)\}= E\{ \frac{d \tau(t)}{dt} \}.
if this relation not correct.
Dear Mohamed Rouamel
Your relation
MR:> "tau (t) is stochastic variable follow Bernoulli distribution if :
\frac{d}{dt} E\{\tau(t)\}= E\{ \frac{d \tau(t)}{dt} \}."
requires severe additional explanation:
1.
Is this a proposal of a definition of a random process following the Bernoulli distribution? or of a theorem?
2.
In each of the cases, one does need to have accepted some definition of the derivative of random process. Do you have some? Let me remind that there are many possibilities. The simplest one is the ordinary derivative of the trajectory; this, however is available for particular random processes only (those, with differentiable trajectories). Another family is formed by processes with finite variation, differentiable as trajectories in the Hilbert spaces, etc.
Best, Joachim Domsta
Thank you very much, Mr. Joachim Domsta and Andrew Paul McKenzie Pegman for your response.
Joachim Domsta, my problem exactly, I have a stochastic variable \tau(t)=\sigma(t)*d(t), where \sigma(t) is a bernouli destrebution and d(t) is continus variable I propose the following relation to get de derivative of \tau(t):
\tau(t) = ( \sigma(t) +\bar \sigma -\bar \sigma)*d(t). then
E\{ \tau(t) \}=E\{ ( \sigma(t) -\bar \sigma)*d(t)+\bar \sigma*d(t)\}.
where \bar \sigma is the expected value of \sigma(t). then calculate the derivative of this last relation as:
\frac{ d E\{ \tau(t) \}}{dt}=\frac{ d}{dt} E\{ ( \sigma(t) -\bar \sigma)*d(t)+\bar \sigma*d(t)\}. then
\frac{ d E\{ \tau(t) \}}{dt}= \bar \sigma* \frac{ d d(t)}{dt}.
with E\{( \sigma(t) -\bar \sigma)\}=0.
but I need the E\{ \frac{ d \tau(t)}{dt} \}. for that, I search if relation
E\{ \{ d \tau(t)}{dt} \} = \frac{ d E\{ \tau(t) \}}{dt} is correct or not.
To Mohamed Rouamel ,
I have cancelled my answer been here, since it was not compatible to the content of your jpg file. Btw, I do not understand your proposal presented wothin the ralation2.png file.
The only idea I can see is as follows:
You need the derivative of the expectation of \tau = \sigma * d where sigma is a process with constant expactation and d is a smooth determininistic psignal.
If this is what you want to have, then you do not need to introduce the derivative of process \tau since then E{\tau(t)} = d(t) for every t. Therefore also the derivative of the expectation equals the derivative of the signal. However, let me stress again, this is not any derivative of the process but of its expectation, only.
Best, JoaD
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