AAPL - Earnings this Monday - Buy OTM QQQ weekly calls on Monday

Quote from lcs:
still require personal judgment on the weights between events vols and everyday vols.
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how is that? event is 1 business day and everything else ambient vol.
 
Not much trading action in the QQQ this week, down only 0.25% since I opened the position on Monday. The Nov1 83.50 weekly calls have expired worthless.



QQQ 5-day chart
qqqq-1.png

Red dot - Bought 10 contracts 83.50 calls at $0.32
 
Quote from sle:

how is that? event is 1 business day and everything else ambient vol.
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Compared to 5D expiry, 1M vols breakdown between event and ambient should place more weight on events vols rather than 1:19 (assuming 20BD). Maybe 2:18, ..., 5:15, ...? depending on the events within the 1M period. Likewise for 2M vs 1M.

Am I misinterpreting something here? thanks.
 
Hey FXforex
here is bullish idea for you I have found, you can also do it in aapl7
its a butterfly; 1 AAPL Dec 525 Call was purchased at 15.00, two 550 Call were sold for 7.00 each and then 1 AAPL 575 Call was bought for 3.00. The net cost for this trade is 4.00, which is also the maximum loss. If AAPL is right at 550.00 at expiration then the result will be a profit of 21.00.
seems attractive to me
 
Quote from lcs:

Compared to 5D expiry, 1M vols breakdown between event and ambient should place more weight on events vols rather than 1:19 (assuming 20BD). Maybe 2:18, ..., 5:15, ...? depending on the events within the 1M period. Likewise for 2M vs 1M.

Am I misinterpreting something here? thanks.
More...
Dude, it's very simple - you take ATM volatility for two expirations, say e1 and e2. This gives you a system of linear equations:

vol(e1)^2 * BD(e1)/252 = ambient_vol^2 * (BD(e1)-1)/252 + event_vol^2 * 1/252
vol(e2)^2 * BD(e2)/252 = ambient_vol^2 * (BD(e2)-1)/252 + event_vol^2 * 1/252

I trust you can solve a system of linear equations? :)
 
Quote from sle:

Dude, it's very simple - you take ATM volatility for two expirations, say e1 and e2. This gives you a system of linear equations:

vol(e1)^2 * BD(e1)/252 = ambient_vol^2 * (BD(e1)-1)/252 + event_vol^2 * 1/252
vol(e2)^2 * BD(e2)/252 = ambient_vol^2 * (BD(e2)-1)/252 + event_vol^2 * 1/252

I trust you can solve a system of linear equations? :)
More...

My question was why is it (BD(e1)-1)/252 : 1/252?
And not (BD(e1)-2)/252 : 2/252 or (BD(e1)-3)/252 : 3/252 or so on?
If your e1 and e2 is large enough, the 2nd term becomes less significant. In this case, the vol term structure would have been flat
thanks!
this addition of variance was extremely helpful. cheers!
 
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