Frequently Asked Questions In Quantitative Finance

Chapter 2: FAQs 63

0.1) is

1

√

2 π 0. 1

exp

(

−

2. 62

2 × 0. 12

)

= 610 −^147.

Very, very unlikely!

(N.B. The word ‘probability’ is in inverted commas to
emphasize the fact that this is the value of the proba-
bility density function, not the actual probability. The
probability of picking exactly−2.6 is, of course, zero.)

The ‘probability’ of picking the number−2.6 from hat B
(having a mean of zero and a standard deviation of 1) is

1

√

2 π 1

exp

(

−

2. 62

2 × 12

)

= 0 .014,

and from hat C (having a mean of zero and a standard
deviation of 10)

1

√

2 π 10

exp

(

−

2. 62

2 × 102

)

= 0. 039.

We would conclude that hat C is the most likely, since it
has the highest probability for picking the number−2.6.

We now pick a second number from the same hat, it is
0.37. This looks more likely to have come from hat B.
We get the following table of probabilities.

Hat −2.6 0.37 Joint

A610−^147 0.004 2 10−^149

B 0.014 0.372 0.005

C 0.039 0.040 0.002

The second column represents the probability of draw-
ing the number−2.6 from each of the hats, the third