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I strongly encourage you to work through the case where $x$ can take two values (e.g. 0 = dominant, 1 = recessive as in your example), as it a useful exercise for understanding the logistic model and also the MLE approach.

The model in question is supposed to be $$ Y_i \sim \mathrm{Bernoulli}(\mathrm{logit}(p_i)), $$ where $p_i = a + bx_i$, $i = 1,\ldots,n$. Suppose the observed values are $y_1, \ldots, y_n$.

Homework:

1. What is the probability of observing this particular string of $y_i$ (assuming they are all independent). Hint: split according to whether $y_i = 0$ or $1$ and similarly for $x_i$.
2. What is the log of this?
3. What happens to the log likelihood if $a$ gets large or small (negative)? Same question for $b$.
4. Conclude that you can maximise the log likelihood by differentiating in $a$ and $b$ then finding where the derivative is zero (separately for $a$ and $b$).

OK, for the final answer: Write $\#_{l,m} = \#\{i : y_i = l, x_i = m\}$, so $\#_{0,1}$ is the number of observations with $y_i = 0$ and $x_i = 1$ etc.

You'll see the problem can be parameterised more easily in terms of $$ p = \frac{e^{a}}{1 + e^{a}} $$ and $$ r = \frac{e^{a+b}}{1 + e^{a+b}}. $$

Assuming I didn't make a mistake, we find $$ r = \frac{\#_{1,1}}{\#_{0,1} + \#_{1,1}}, $$ i.e. the fraction of cases with $y_i = 1$ when the independent variable has $x_i = 1$.

Then, $$ p = \frac{\#_{1,0}}{\#_{0,0} + \#_{1,0}} $$ is the fraction when $x_i = 0$.

So, $a + b = \ln(r/(1-r))$ is the log odds for $y_i = 1$ when $x_i = 1$.

Finally, $a = \ln(p/(1-p))$ and $b = \ln(r(1-p)/p(1-r))$ is the log odds ratio.

I strongly encourage you to work through the case where $x$ can take two values (e.g. 0 = dominant, 1 = recessive as in your example), as it a useful exercise for understanding the logistic model and also the MLE approach.

The model in question is supposed to be $$ Y_i \sim \mathrm{Bernoulli}(p_i), $$ where $p_i = \mathrm{logistic}(a + bx_i)$, $i = 1,\ldots,n$. Suppose the observed values are $y_1, \ldots, y_n$.

Homework:

1. What is the probability of observing this particular string of $y_i$ (assuming they are all independent). Hint: split according to whether $y_i = 0$ or $1$ and similarly for $x_i$.
2. What is the log of this?
3. What happens to the log likelihood if $a$ gets large or small (negative)? Same question for $b$.
4. Conclude that you can maximise the log likelihood by differentiating in $a$ and $b$ then finding where the derivative is zero (separately for $a$ and $b$).

OK, for the final answer: Write $\#_{l,m} = \#\{i : y_i = l, x_i = m\}$, so $\#_{0,1}$ is the number of observations with $y_i = 0$ and $x_i = 1$ etc.

You'll see the problem can be parameterised more easily in terms of $$ p = \frac{e^{a}}{1 + e^{a}} $$ and $$ r = \frac{e^{a+b}}{1 + e^{a+b}}. $$

Assuming I didn't make a mistake, we find $$ r = \frac{\#_{1,1}}{\#_{0,1} + \#_{1,1}}, $$ i.e. the fraction of cases with $y_i = 1$ when the independent variable has $x_i = 1$.

Then, $$ p = \frac{\#_{1,0}}{\#_{0,0} + \#_{1,0}} $$ is the fraction when $x_i = 0$.

So, $a + b = \ln(r/(1-r))$ is the log odds for $y_i = 1$ when $x_i = 1$.

Finally, $a = \ln(p/(1-p))$ and $b = \ln(r(1-p)/p(1-r))$ is the log odds ratio.

I strongly encourage you to work through the case where $x$ can take two values (e.g. 0 = dominant, 1 = recessive as in your example), as it a useful exercise for understanding the logistic model and also the MLE approach.

The model in question is supposed to be $$ Y_i \sim \mathrm{Bernoulli}(\mathrm{logit}(p_i)), $$ where $p_i = a + bx_i$, $i = 1,\ldots,n$. Suppose the observed values are $y_1, \ldots, y_n$.

Homework:

1. What is the probability of observing this particular string of $y_i$ (assuming they are all independent). Hint: split according to whether $y_i = 0$ or $1$ and similarly for $x_i$.
2. What is the log of this?
3. What happens to the log likelihood if $a$ gets large or small (negative)? Same question for $b$.
4. Conclude that you can maximise the log likelihood by differentiating in $a$ and $b$ then finding where the derivative is zero (separately for $a$ and $b$).

OK, for the final answer: Write $\#_{l,m} = \#\{i : y_i = l, x_i = m\}$, so $\#_{0,1}$ is the number of observations with $y_i = 0$ and $x_i = 1$ etc.

You'll see the problem can be parameterised more easily in terms of $$ p = \frac{e^{a}}{1 + e^{a}} $$ and $$ r = \frac{e^{a+b}}{1 + e^{a+b}}. $$

Assuming I didn't make a mistake, we find $$ r = \frac{\#_{1,1}}{\#_{0,1} + \#_{1,1}}, $$ i.e. the fraction of cases with $x_i = 1$ that have $y_i = 1$.

In terms of $r$, we get $$ p = \frac{\#_{1,0}}{\#_{0,0} + \#_{1,0}}. $$

So, $a + b = \ln(r/(1-r))$ is the log odds have $y_i = 1$ when $x_i = 1$.

Finally, $a = \ln(p/(1-p))$ and $b = \ln(r(1-p)/p(1-r))$ is the log odds ratio.

I strongly encourage you to work through the case where $x$ can take two values (e.g. 0 = dominant, 1 = recessive as in your example), as it a useful exercise for understanding the logistic model and also the MLE approach.

The model in question is supposed to be $$ Y_i \sim \mathrm{Bernoulli}(\mathrm{logit}(p_i)), $$ where $p_i = a + bx_i$, $i = 1,\ldots,n$. Suppose the observed values are $y_1, \ldots, y_n$.

Homework:

1. What is the probability of observing this particular string of $y_i$ (assuming they are all independent). Hint: split according to whether $y_i = 0$ or $1$ and similarly for $x_i$.
2. What is the log of this?
3. What happens to the log likelihood if $a$ gets large or small (negative)? Same question for $b$.
4. Conclude that you can maximise the log likelihood by differentiating in $a$ and $b$ then finding where the derivative is zero (separately for $a$ and $b$).

OK, for the final answer: Write $\#_{l,m} = \#\{i : y_i = l, x_i = m\}$, so $\#_{0,1}$ is the number of observations with $y_i = 0$ and $x_i = 1$ etc.

You'll see the problem can be parameterised more easily in terms of $$ p = \frac{e^{a}}{1 + e^{a}} $$ and $$ r = \frac{e^{a+b}}{1 + e^{a+b}}. $$

Assuming I didn't make a mistake, we find $$ r = \frac{\#_{1,1}}{\#_{0,1} + \#_{1,1}}, $$ i.e. the fraction of cases with $y_i = 1$ when the independent variable has $x_i = 1$.

Then, $$ p = \frac{\#_{1,0}}{\#_{0,0} + \#_{1,0}} $$ is the fraction when $x_i = 0$.

So, $a + b = \ln(r/(1-r))$ is the log odds for $y_i = 1$ when $x_i = 1$.

Finally, $a = \ln(p/(1-p))$ and $b = \ln(r(1-p)/p(1-r))$ is the log odds ratio.

I strongly encourage you to work through the case where $x$ can take two values (e.g. 0 = dominant, 1 = recessive as in your example), as it a useful exercise for understanding the logistic model and also the MLE approach.

The model in question is supposed to be $$ Y_i \sim \mathrm{Bernoulli}(\mathrm{logit}(p_i)), $$ where $p_i = a + bx_i$, $i = 1,\ldots,n$. Suppose the observed values are $y_1, \ldots, y_n$.

Homework:

1. What is the probability of observing this particular string of $y_i$ (assuming they are all independent). Hint: split according to whether $y_i = 0$ or $1$ and similarly for $x_i$.
2. What is the log of this?
3. What happens to the log likelihood if $a$ gets large or small (negative)? Same question for $b$.
4. Conclude that you can maximise the log likelihood by differentiating in $a$ and $b$ then finding where the derivative is zero (separately for $a$ and $b$).

OK, for the final answer: Write $\#_{l,m} = \#\{i : y_i = l, x_i = m\}$, so $\#_{0,1}$ is the number of observations with $y_i = 0$ and $x_i = 1$ etc.

You'll see the problem can be parameterised more easily in terms of $$ p = \frac{e^{a}}{1 + e^{a}} $$ and $$ r = \frac{e^{a+b}}{1 + e^{a+b}}. $$

Assuming I didn't make a mistake, we find $$ r = \frac{\#_{1,1}}{\#_{0,1} + \#_{1,1}}, $$ i.e. the fraction of cases with $x_i = 1$ that have $y_i = 1$.

In terms of $r$, we get $$ p(\#_{0,0} + \#_{1,0} ) = \#_{1,0} + \#_{1,1} + r(\#_{0,1} - \#_{0,0}) $$

So, $a + b = \ln(r/(1-r))$ is the log odds have $y_i = 1$ when $x_i = 1$.

Finally, $a = \ln(p/(1-p))$ and $b = \ln(r(1-p)/p(1-r))$.

I strongly encourage you to work through the case where $x$ can take two values (e.g. 0 = dominant, 1 = recessive as in your example), as it a useful exercise for understanding the logistic model and also the MLE approach.

The model in question is supposed to be $$ Y_i \sim \mathrm{Bernoulli}(\mathrm{logit}(p_i)), $$ where $p_i = a + bx_i$, $i = 1,\ldots,n$. Suppose the observed values are $y_1, \ldots, y_n$.

Homework:

1. What is the probability of observing this particular string of $y_i$ (assuming they are all independent). Hint: split according to whether $y_i = 0$ or $1$ and similarly for $x_i$.
2. What is the log of this?
3. What happens to the log likelihood if $a$ gets large or small (negative)? Same question for $b$.
4. Conclude that you can maximise the log likelihood by differentiating in $a$ and $b$ then finding where the derivative is zero (separately for $a$ and $b$).

OK, for the final answer: Write $\#_{l,m} = \#\{i : y_i = l, x_i = m\}$, so $\#_{0,1}$ is the number of observations with $y_i = 0$ and $x_i = 1$ etc.

You'll see the problem can be parameterised more easily in terms of $$ p = \frac{e^{a}}{1 + e^{a}} $$ and $$ r = \frac{e^{a+b}}{1 + e^{a+b}}. $$

Assuming I didn't make a mistake, we find $$ r = \frac{\#_{1,1}}{\#_{0,1} + \#_{1,1}}, $$ i.e. the fraction of cases with $x_i = 1$ that have $y_i = 1$.

In terms of $r$, we get $$ p = \frac{\#_{1,0}}{\#_{0,0} + \#_{1,0}}. $$

So, $a + b = \ln(r/(1-r))$ is the log odds have $y_i = 1$ when $x_i = 1$.

Finally, $a = \ln(p/(1-p))$ and $b = \ln(r(1-p)/p(1-r))$ is the log odds ratio.