1

Struggling a little bit with this concept in class.

So using R I trialled 10,000 students taking a test (10 question, 4 multiple choice answers, only one of which is correct). I have a table of frequencies 0-10 how many correct answers each student got based on pure guessing.

Now I need to calculate E(X|X>5) i.e. the mean of passing scores. I'm not really sure how to approach this and not sure how to apply conditional expectations - whenever I look at examples they seem to use two random variable and not one? I think my understanding may be lacking.

Looking for simple and straight forward answers.

Edit. Is there a way to answer this question with Bayes theorem? this is a homework assignment and it was heavily implied that this is an exercise of conditional expectation.

kitkat1224
  • 45

1 Answers1

2

If $X$ is a (suitable) random variable with CDF $\mathsf F_X$ then: $$\mathsf E[X\mid X>5]=\frac{\mathsf EX\mathbf1_{X>5}}{\mathsf P(X>5)}=\frac{\int_5^{\infty}xd\mathsf F_X(x)}{\int_5^{\infty}d\mathsf F_X(x)}$$

drhab
  • 151,093
  • The distribution is discrete not continuous. – kitkat1224 May 01 '18 at 09:55
  • 1
    What you meet in my answer is valid for continuous and discrete. For discrete with PDF $p$ then the numerator equals $\sum_{k>5}xp(x)$ and the denominator equals $\sum_{k>5}p(x)$. – drhab May 01 '18 at 09:58
  • In my former comment I abusively said "PDF" (probability density function). It should be "PMF" (probability mass function). – drhab May 01 '18 at 11:57