To calculate the first derivative of \(\ell(\beta)\), we need to differentiate each term in the summation with respect to \(\beta\) and sum them up:
\[\ell'(\beta) = \sum_{i=1}^{n} \left(Y_i X_i - \frac{X_i e^{\beta X_i}}{1 + e^{\beta X_i}}\right)\]
To calculate the second derivative of \(\ell(\beta)\), we need to differentiate each term in the first derivative with respect to \(\beta\) and sum them up:
\[\ell''(\beta) = \sum_{i=1}^{n} \left(-\frac{X_i^2 e^{\beta X_i}}{(1 + e^{\beta X_i})^2} - \frac{X_i^2 e^{\beta X_i}}{1 + e^{\beta X_i}} + \frac{X_i^2 e^{2\beta X_i}}{(1 + e^{\beta X_i})^2}\right)\]
From the first derivative, we can conclude that \(\ell'(\beta)\) represents the gradient or the direction of steepest ascent of the log-likelihood function \(\ell(\beta)\). This means that it tells us in which direction \(\beta\) should be adjusted to maximize \(\ell(\beta)\).
Note: This is not financial advice, and should not be taken as such.
We consider a 1-dimensional logistic regression problem, i.e., assume that data \, X_ i \in \mathbb {R}, i = 1, \dots , n \, is given and that get independent observations of
Y_ i | X_ i \sim \textsf{Ber}\left( \frac{\mathbf e^{\beta X_ i}}{1 + \mathbf e^{\beta X_ i}} \right),
where \, \beta \in \mathbb {R} \,.
Moreover, recall that the associated log likelihood for \, \beta \, is then given by
\ell (\beta ) = \sum _{i=1}^{n} \left( Y_ i X_ i \beta - \ln (1 + \exp (X_ i \beta )) \right)
Calculate the first and second derivate of \, \ell \,. Instructions: The summation \sum _{i=1}^{n} is already placed to the left of the answer box. Enter the summands in terms of \beta, X_ i (enter “X_i") and Y_ i (enter “Y_i").
\displaystyle \ell '(\beta ) = \sum _{i=1}^{n}
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\displaystyle \ell ^{\prime \prime }(\beta ) = \sum _{i=1}^{n}
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What can you conclude about \, \ell '(\beta ) \,?
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