# OAS Mock Schewser Mock Question Confusion

I’m having trouble understanding this question - Schweser Vol 2 Mock 3 in Fixed Income:

Question: If OAS of a straight is 48bps which of the following is most accurate?

A) OAS of callable bond will be greater than 48bps and OAS of conv. bonds will be less than 48bps

B) OAS of convertible bonds will be less than 48bps while OAS of putable will be greater than 48bps

C) OAS of callable putable and convertible bond should be equal to 48bps

I answered ‘B’ based on the fact that for straight bond OAS = Z-Spread.

thus OAS for putable = Z + \$Putable

OAS for conv = Z - \$Convertible

Can someone help clarify?

OAS removes the value of the options.

That’s correct…confused as to why though.

So is the idea that in any scenario in our tree i.e increasing/decreasing volatility once we remove the option value, the OAS = OAS of an option free (Z-spread)

okay… imma take a go at this:

For example: Vcallable = Vstraight - Vcall option

So when you value the callable bond, you first find what the bond is worth at each node, adjust for the fact that there is an option (so if call price is \$100, and bond is 102, that value at the node would be \$100). So what you have essentially done after this adjustment at each node is removed the call option from this callable bond so that:

Vcallable = Vstraight

OAS is calculated after you adjust for the option. Therefore --> OAS --> ‘removes’ the value of the options

If OAS for your straight is 48bps then the OAS for your callable must also be 48bps