Probability/Statistics (really lost)

[HoldenJ]

Srry there were a little typing error. It should be: b)
Mean: E(X)=0,267*1+0,131*2+0,053*3+0,014*4+0,007*5*0,0*6=0,779

But the Mean is still the same.

It will be Chebyshev's inequality. Which tell me how many % of the observations will lay in a range of the StdDev from mean.

Where I am lost with my values is:
The total amount of observations is 100%, where the mean is 77,9%, and then the StdDev is 102% according to my calculations, which could be wrong.

The StdDev should not be that large believe I.

 

[D H]

Srry there were a little typing error. It should be: b)
Mean: E(X)=0,267*1+0,131*2+0,053*3+0,014*4+0,007*5*0,0*6=0,779

Another typo! (You have "0,007*5*0,0*6"). But you're correct; it's just a typo. You do have the correct mean and the correct standard deviation.

Where I am lost with my values is:
The total amount of observations is 100%, where the mean is 77,9%, and then the StdDev is 102% according to my calculations, which could be wrong.

The StdDev should not be that large believe I.

That you said the mean is 77,9% and the standard deviation is 102% means you may have a fundamental misunderstanding of those concepts. They are not percentages. The mean is 0.779 days, not 77.9%, and the standard deviation is 1.0296 days, not 102.96%. Do not convert these into percentages. It doesn't make sense.

Does this help you understand how to apply Chebychev's inequality?

 

[D H]

Ah, thanks for clearing that up for me, D H.

Now to finding event B:

for n=1, event A is the P(N≥1)=47,2
and event B is the probability for the following day, P(N≥2)=20,5

for n=2, event A is the P(N≥1)=20,5
and event B is the probability for the following day, P(N≥2)=7,4

...and so on up to n=4.

Gone off track again? And the intersection: If it 1-20,5+7,4?
I still fail to wrap my head around the Venn diagram.

You have the correct event B.

You do not have the intersection correct. Think back to why P(N≥3) is larger than P(N=3). It contains P(N=3) and P(N=4) and P(N=5). Your Venn diagram for P(N≥3) should encompass those three events.

Trick question: Given a set C and some other set D that is a subset of set C, what is the intersection of sets C and D?

Now apply this to your events A and B.

 

[HoldenJ]

Havent sleept for 2 days in a row, so srry for the typos..

Thank you very much. it gives totally sense now. I don't know why I thought it should be percentages..

Much appreciated for your fine help.

 

[anonymousk]

so 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 26,7?

and for the next A: P(N≥2), and B: P(N≥3), the intersection is 13,1?

answer to trick question: to me it somehow seems logical subtracting D from C to get the intersection.

 

[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.