A quick question from a newbie about Delta

[Garbanzo]

Hey there,am new around here

I've been reading a book on option's trading and the author gives an example of how to substitute a long stock position with a deep in the money option that has a delta of about 100%.
The cost of such an option according to the book's example for 100 MSFT shares is at 1200$ for a 12$strike price while MS was trading at 22$ or so.
The author also have another example of why it would be detrimental of buy an out of the money option contract,one with a strike of 27$ has a delta of just 22% but the cost is only 40$ for the contract.
So my question is this,assuming I reach the decision that MS was gonna go sky-high and reach 27 or beyond,how about instead of following the author's advice I just bought just instead of 1 27$ call contract with a delta of 22% but 5?
This way,at a cost of 5*40= 200$ I basically control the same position as the example mentioned by the author right?
I get almost the exact same upward potential as the 12$ option?



Thanks in advance!

 

[Martinghoul]

Can you reformulate your question without referencing the various unknown statements in the book? I kinda understand the question, I think, but I believe everyone, incl yourself, would benefit from clarifying it. While you're doing it, think about probability and you may answer your own questions.

 

[ns1000]

That's the effect of leverage. Based on your figures, you can buy 6x as many $27 calls as $12 for the same price. If the stock ramps, this is all fine, but if it shoots up to $26.99 you lose all your money (assuming you hold until expiry). Buying something so far out of the money is like putting everything into a penny share - if it goes right you're quids in, but otherwise you're broke. The upward potential is only the same once the stock hits $27. My advice would be to use the out-of the money option as part of a strategy rather than an end to themselves.

 

[Garbanzo]

That's the effect of leverage. Based on your figures, you can buy 6x as many $27 calls as $12 for the same price. If the stock ramps, this is all fine, but if it shoots up to $26.99 you lose all your money (assuming you hold until expiry). Buying something so far out of the money is like putting everything into a penny share - if it goes right you're quids in, but otherwise you're broke. The upward potential is only the same once the stock hits $27. My advice would be to use the out-of the money option as part of a strategy rather than an end to themselves.

Yeah I understand that but assuming that Delta is 22%, doesn't that mean that for each 1$ the stock moves up the option should gain 22 cents?
And by buying 5 or 6 such deep out of the money options,am I not pretty much getting the same position as buying that much more expensive deep in the money option with 100% delta?

 

[Garbanzo]

Can you reformulate your question without referencing the various unknown statements in the book? I kinda understand the question, I think, but I believe everyone, incl yourself, would benefit from clarifying it. While you're doing it, think about probability and you may answer your own questions.

I am sorry if my question confused you.
Basically he is trying to show the reader how to replicate a long stock position via a deep in the money call option. The deep in the money call option has a Delta of 100%.
The deep out of the money has a delta of 22%.
My question is,can I replicate that deep in the money option just by buying 5 deep out of the money options since their deltas add up and in total they are just a fraction of the cost of the deep in the money?


Thank you

 

[ns1000]

Yeah I understand that but assuming that Delta is 22%, doesn't that mean that for each 1$ the stock moves up the option should gain 22 cents?
And by buying 5 or 6 such deep out of the money options,am I not pretty much getting the same position as buying that much more expensive deep in the money option with 100% delta?

That will work only as long as the delta remains the same. If the stock moves up quickly, the OOM options will get closer to the money so the delta will increase and your synthetic position will be commensurately larger. If the stock doesn't climb, the delta will shrink as time passes. The point of buying a heavily ITM option is that it's unlikely to be affected so quickly (although it will be affected) by theta decay. It's delta is going to be as near as damn it 100% for a lot longer than your OOM calls is 22% if the price doesn't move much.

 

[Garbanzo]

That will work only as long as the delta remains the same. If the stock moves up quickly, the OOM options will get closer to the money so the delta will increase and your synthetic position will be commensurately larger. If the stock doesn't climb, the delta will shrink as time passes. The point of buying a heavily ITM option is that it's unlikely to be affected so quickly (although it will be affected) by theta decay. It's delta is going to be as near as damn it 100% for a lot longer than your OOM calls is 22% if the price doesn't move much.

Yeah that makes sense,its the risk you take from time decay for being cheap
But I guess for short term trades that you expect to materialize in a matter of weeks you could always just buy OOM options with 3 months left and still be fine.

Do you happen to know of any way to measure the effect of time decay on delta?

 

[Martinghoul]

I am sorry if my question confused you.
Basically he is trying to show the reader how to replicate a long stock position via a deep in the money call option. The deep in the money call option has a Delta of 100%.
The deep out of the money has a delta of 22%.
My question is,can I replicate that deep in the money option just by buying 5 deep out of the money options since their deltas add up and in total they are just a fraction of the cost of the deep in the money?
Thank you

You can replicate anything you like, but you need to make a distinction between the two alternative positions. Intuitively, deltas being equal means that in both cases payout * Prob(payout recvd) is the same. However, that doesn't mean that elements of the multiplication are the same. Don't know if it helps, 'cause I might not be answering your question.