Probability/Statistics (really lost)

[D H]

The trick question is very pertinent to this problem, anonymousk.

The intersection of two sets is the subset of each set that is common to both sets. Subtracting set D from set C yields the part of set C that does not contain any of D. That is not the intersection of C and D.

Try drawing a sequence of pictures. Represent the two sets (C and D) as circles, with the circle representing set C larger than that representing set D. Start the sequence with the circles barely intersecting. That little lens-shaped object that is inside both circles -- that's the intersection of the two sets. Now draw it again, but this time with a more significant intersection. And again, with an even greater intersection. And one more time, where D is completely inside C. What's the intersection look like now?

 

[anonymousk]

The lens shape you're talking about, isn't that the union? Where the two circles overlap each other.
By trying the Venn diagram again, I came to the conclusion that the intersection of set C and D (Where D is a subset of C) is D. Hmm.

 

[D H]

 

[anonymousk]

for n=1

Event A: P(N≥1)=47,2
Event B: P(N≥2)=20,5

P(N≥1) contains (26,7)+(13,1)+(5,3)+(1,4)+(0,7)
P(N≥2) contains (13,1)+(5,3)+(1,4)+(0,7)

So the intersection in this case is (13,1)+(5,3)+(1,4)+(0,7)=20,5 , which is the same value as event B (in this case)?
$P(B|A) = \frac{P(A\cap B)}{P(A)}$.

$P(B|A) = \frac{20,5}{47,2}$.= 0,434

 

[D H]

Yes! Finally!

The intersection of a set C and a subset D of set C is *always* equal to D.

 

[anonymousk]

Haha, yeah, finally indeed! I have a feeling I won't be having much problems in these assignments again. (If I do, I know where to find you ;) )

$P(B|A) = \frac{P(A\cap B)}{P(A)}$.

$P(B|A) = \frac{20,5}{47,2}$.= 0,4343

$P(B|A) = \frac{7,4}{20,5}$.= 0,3610

$P(B|A) = \frac{2,1}{7,4}$.= 0,2838

$P(B|A) = \frac{0,7}{2,1}$.= 0,3333

Then I'm guessing I multiply these probabilities 0,4343*0,3610*0,2838*0,3333 = 0,0148.

Is this correctly done?

$P(B|A) = \frac{P(A\cap B)}{P(A)}$ = $P(B|A) = \frac{(20,5*7,4*2,1*0,7)}{(47,2*20,5*7,4*2,1)}$= 0,0148 = 1,48%

 

[D H]

There's no need to multiply those probabilities.

 

[anonymousk]

Oh. I'll let them stand separate then. Thanks for your help! :)

 

[anonymousk]

There's actually one last question in the assignment in d) which I didn't include."Estimate based on the calculated probabilities of prior day equity returns contain useful information for predicting the next day's return by investing in the stock."

Not sure how to see this. The probability goes down in d) n=1-3, but 4 it rises again. Not sure how to understand this.

 

[D H]

One way to explain that apparent discrepancy is rounding errors. Probabilities are reported to the tenth of a percent. This rounding will make your calculated conditional probabilities be a bit erroneous, particularly when probabilities are small. For example, P(N=5)=0.7% and P(N=4)=1.4% are small probabilities. That apparent rise at n=4 might be a result of rounding rather than a real rise.