probability example

mouse123

from book 1 CFAI 2009 page 328....

does anyone guide me how did
P(order 1 executes\order 2 executes)=1

in example 3

evolsteevol

This is a conditional probability issue. You should read "P(order 1 executes|order 2 executes) = 1" as the probability that order 1 executes given that you 'know for sure' that order 2 executes.

Now by common sense, and also the CFAI writes it there, "it is certain" that to satisfy the limit order condition of order 2, the price has to pass through the limit order condition of order 1. For these limit buy orders to execute, the price must keep falling from somewhere above $10 and through $9.75.

This essentially means that the conditional probability of order 1 executing, given you know that order 2 executes = 1.

"It is certain" implies probability is 1.

Hope it helps.

scr879

Hey there!

Suppose the market price is currently $12. Limit buy order are always placed below the current market price.

I place two limit buy orders- one at $10 and the second one at $9.75. The probability that the first order will be executed given that the second one has been executed is 1. The second limit order would require the price to fall to $9.75. On its way down from $12 to $9.75, the market price would definitely cross $10, which is the price at which the first order would execute.

Therefore is a certainty (probability = 1) that the buy order at $10 will be executed given that the price has fallen to $9.75 (the price at which the second order was executed).

JGovender

This is actually a over-simplified example. It is not necessary that a limit buy order of 10 must get executed before a limit buy order of 9.75.

Execution of the buy orders depend on many factors - trading volume, blocks of shares, minimum trade lots etc. For instance a matching trade must be found for the limit buy order of 10. If your limit buy order of 10 was for a 10000 shares and the limit buy order of 9.75 was for 1 share and only 1 share was offered in the market, naturally the 10000 share limit order for 10 will not get executed where as the 9.75 will !!!!!

I have over-simiplified as well - naturally we dont expect the trading volume to be 1 for any real stock. There are many other factors as well. The stock exchanges have complex algorithms to match trades.

That being said, from a illustration of probability concepts perspective the CFAI textbook has gone with the assumption that a $ 10 limit order will get executed before a $ 9.75 just because if you are willing to buy a stock at $ 10 you will get the stock first rather than another person willing to buy the same stock at $ 9.75.

Note that the stock exchange works on the basis of bids and offers. You offer a higher price you get the stock first !!! Hence from a pure price perspective, you will definitely get the stock if you offer $10 for it if anyone offering 9.75 gets it. And prob of a certain event is 1.

Good Luck with L1

- VR

Ionutzakis

I completely agree with VR. This is what kept annoying me about this question. I don't see how you can make the assumption with 100% guarantee that the higher price buy-order will be executed. Thus, I was looking it from purely the probability factors; the way I would do in other questions. But, perhaps I'm missing something here, these questions do give me a hard time occasionally. Tree diagrams are the only way I have a shot at answering any of these in the exam.