Score of an Exam (Variance/Measure Problem) I want to explain in accurate way the next problem to a student, using intuiton the student got it the answer, but he wants learn the precise and formal procedure.
The problem is the following:
A math teacher decides to give a multiple-choice test in a measurement theory course. The professor decides to give a penalty for answering a question wrong, explaining that if they do not answer the question they will not have any penalty. The penalty for wrongly answering a question with $n$ options is $\frac{1}{(n-1)}$ multiplied by the value of the question and there is only one correct option. Thus, for example, a true or false question worth 2 points will give a penalty of 2 points for incorrect answers and no penalty for no answer. The teacher justifies his choice by explaining to the students that, with this way of grading, randomly answering a question does not change the average value of their score but does increase its variance. The professor's assertion is True or False?
 A: TLDR: The professors assertion is True
The expectation value can be found by multiplying the probability of a correct/incorrect answer and the pointgain/penalty for that answer, $E[X]=P(X=correct)*pointgain-P(X=wrong)*penalty$. The variance is found as $Var[X]=E[X^2]-(E[X])^2$. In the below scenarios it is assumed that the answer is selected randomly, that the points you get for a correct answer are $w$, and that blank answers neither earn or lose you points.
1 Penalty of $\mathbf{\frac{w}{n-1}}$ for incorrect answers:
The expectation is:
$$
\begin{aligned}
E[X]&=P(correct)w-P(incorrect)\left(\frac{w}{n-1}\right)\\
&=\left(\frac{1}{n}\right)w-\left(\frac{n-1}{n}\right)\left(\frac{w}{n-1}\right)\\
&=0,
\end{aligned}
$$
where $w$ is the weight of each question. The expectation value is zero (average value doesn't change).
The variance is:
$$
\begin{aligned}
Var[X]&=E[X^2]-(E[X])^2\\
&=\left[\left(\frac{1}{n}\right)w^2-\left(\frac{n-1}{n}\right)\left(\frac{w}{n-1}\right)^2\right]-\left[0\right]^2\\
&=\frac{w^2}{n}\left(\frac{n-2}{n-1}\right)
\end{aligned}
$$
which is not zero.
2 Not randomly answering questions you don't know:
The expectation and variance are both zero
$$
\begin{aligned}
E[X]&=0,\\
Var[X]&=0.
\end{aligned}
$$
3 No penalty for wrong questions:
Assuming the answer is selected randomly. The expectation would be:
$$
\begin{aligned}
E[X]&=P(X=correct)w\\
&=\frac{w}{n}\neq 0,
\end{aligned}
$$
and the variance would be:
$$
\begin{aligned}
Var[X]&=\left[\frac{w^2}{n}\right]-\left[\frac{w}{n}\right]^2\\
&=\frac{w^2\left(n-1\right)}{n^2}
\end{aligned}
$$
4 Penalty of $\mathbf{w}$ for incorrect answers:
Assuming the answer is selected randomly. The expectation would be:
$$
\begin{aligned}
E[X]&=\left(\frac{1}{n}\right)w-\left(\frac{n-1}{n}\right)w\\
&=\left(\frac{2-n}{n}\right)w,
\end{aligned}
$$
In this case if you didn't know the answer and there are more than 2 options for the questions you don't want to randomly answer questions.
and the variance would be:
$$
\begin{aligned}
Var[X]&=\left[\left(\frac{1}{n}\right)w^2-\left(\frac{n-1}{n}\right)w^2\right]-\left[\left(\frac{2-n}{n}\right)w\right]^2\\
&=2w^2\left(\frac{2-n}{n^2}\right)\left(n-1\right)
\end{aligned}
$$
Now if you were to take a sum over the number of questions that you guess and add the expectation and variance to scores where the answers are known, the total expectation and variance could be determined. If you do this you will find that the professor's assertion is True. The average value of their score is not affected by randomly guessing incorrect answers, while the variance does increase.
