# Sector rotation question

Hi folks,

Here is the question:

And here are the given answers:

Question:
2) Where is the 0.05 and -0.05 in the yellow highlighted portion from? The only 0.05 I see so far is the monthly active risk, but that is not linked to returns, right?

1. Can someone explain this statement to me? “If active risk is limited to 5.20%, the deviation from the benchmark weights of 65% and 35% is limited to (5.20% / 17.32%) = 30%.” I do not understand the concept behind this.

That’s exactly where it comes from: \sigma_{C}

In the answer to 2, you have a formula
annualized active return = IC \times \sqrt{\textrm{BR}}\times \sigma_{A}
where \sigma_{A} is the annualized active risk,
so yes, active risk is linked to active returns.

The expression you have highlighted in yellow
(0.60)(0.50)+(0.40)(-0.50)=(0.60-0.40)(0.50)=IC \times\sigma_{C}

3: The active risk if the deviation from the benchmark is 100% is 17.32%
I assume that 100% deviation from the benchmark means either
staples 165(=65+100) and discretionary -65(=35-100) or
staples -35(=65-100) and discretionary 135(=35+100)

The active risk is proportional to the deviation from the benchmark:
If the deviation from the benchmark is x\% then the active risk will be (x/100)\times 17.32\%

You need to find the value of x such that (x/100)\times 17.32\% =5.32\% ,
so x=100\times 5.2/17.32=30\%

Thanks @guest

1. Doesn’t the answer then assume that TC = 1? Why are we able to make this assumption?

2. Thank you so much. This is super clear

1 Like

I think you need the help of an expert like S2000magician or Mikey here

As you say, the formula in the answer they give to q2
annualized active return = IC \times \sqrt{\textrm{BR}}\times \sigma_{A}
should be
annualized active return = TC \times IC \times \sqrt{\textrm{BR}}\times \sigma_{A}
where TC is the Transfer Coefficient
However, when there are no constraints, the Transfer Coefficient TC =1, so for 2., the formula
reduces to the formula they gave.

For 3., there is a constraint (the active risk can’t exceed 5.32%), so the transfer coefficient will be <1.
I’ve got a suspicion that TC =0.3 (from the 30%) for 3. Hopefully one of the experts can either confirm this or set us straight,