September 2006

[Louisa]

Okay, two things have been bugging me about this one. Anyone want to discuss?
1) So they decided to test who was reading the whole question with the plots of investment returns. If you didn't spot the third plot coming halfway down the page, you could go over the edge of the paper like me. Do you think I would have been allowed to sellotape more graph paper on? A bit primary school...
2) What on earth does "show that the optimum is a minimum" mean?

Actually, come to think of it, does anyone know whether it is permitted to draw graphs in pencil in exams?

 

[StevieG4captain]

I didn't take it this session

(am planning to sit it next April)

I assume the answer to question 2 would involve calculating the 2nd derivative of whatever variable you were being asked to find the optimum value for. If you were then able to show that this is always positive over the whole of the domain, you'd effectively be proving that the optimum value is a minimum.

Taking a step back, to show that the value was indeed optimum, you would have needed to have differentiated it once (wrt the variable in question), set this to 0 and solved for the variable of interest.

Again, like I said, haven't actually seen the question, so I'm making a lot of assumptions based on your post. If I've got the wrong end of the stick, apologies.

 

[Rob]

Ct3 2006

Okay, two things have been bugging me about this one. Anyone want to discuss?

...

Actually, come to think of it, does anyone know whether it is permitted to draw graphs in pencil in exams?

I wondered about this, given that it explicitly states that answers must not be written in pencil. It seems kind of wrong to draw a graph in pen but I imagine it's more for security than anything else.

Did anyone think that 3 marks seemed rather a lot for question 2: "If A and B are independent show that P(A) and P(¬B) are independent (¬B = complement of B)"? Is the answer not just:
If A and ¬B are independent then P(A|¬B) = P(A|¬¬B) = P(A), but P(A|¬¬B) = P(A|B) and we know that P(A|B) = P(A) since A & B are independent. Hence A & ¬B are independent.

The phrase gift horse and mouth springs to mind, if only I hadn't completely overcomplicated things and confuddled myself

 

[Louisa]

Excellent deduction StevieG. Yes it was an minimisation-type question. The first clause was "find the value of ? to minimise ?" and the second "show that the optimum is a minimum".
You could well be right about what they were after - I'm probably okay then, I always make a point of showing that minimums are minimums. Nice of them to try to give a hint, but pretty weird question-formulation IMO.

 

[Louisa]

Did anyone think that 3 marks seemed rather a lot for question 2: "If A and B are independent show that P(A) and P(¬B) are independent (¬B = complement of B)"? Is the answer not just:
If A and ¬B are independent then P(A|¬B) = P(A|¬¬B) = P(A), but P(A|¬¬B) = P(A|B) and we know that P(A|B) = P(A) since A & B are independent. Hence A & ¬B are independent.

Looks like you're using a different definition of independence to me - yes, that one does make it a bit of a gimme! Wish I'd thought of that. I used P(A cap B)=P(A)P(B), which gives it a bit more substance, though still not a lot.

 

[Rob]

Brainache

Looks like you're using a different definition of independence to me - yes, that one does make it a bit of a gimme! Wish I'd thought of that. I used P(A cap B)=P(A)P(B), which gives it a bit more substance, though still not a lot.

Maybe I'm still having brainache from the exam but how did you show it using P(A ^ B)?

 

[Louisa]

P(A ^ ¬B) = P(A - (A ^ B)) = P(A) - P(A ^ B) etc.

Talk about brainache - I've been so spaced out between exams. And in exams on occasion :-(