Probability/Statistics (really lost)

[anonymousk]

I'm not a native English speaker, and this assignment wasn't originally in English, so I had to translate it to English, thus the grammar errors, but hopefully it's understandable)
-----------------------------------
For an investment in an equity assessed in general to be the following probabilities attached to the exact number of days in a row with positive returns

One day with a positive return , 2 day negative return 26.7%

2 consecutive days with positive returns , 3. day negative return……………. 13.1%

3 consecutive days with positive returns , 4. day negative return……………. 5.3%

4 consecutive days with positive returns , 5. day negative return …………….1.4%

5 consecutive days with positive returns , 6. day negative return …………….0.7%

6 or more consecutive days with positive returns …………………………………….0.0%

For example , there seem to be a probability of 26.7% on any given workday to achieve a positive return and then on the following day to achieve a negative return. Similarly, there seem to be a probability of 13.1% on any given workday to achieve a positive return and then on the following day also to achieve a positive return while on the third day to achieve a negative return.

a) Calculate by using the table above the probability of maximum n days in a row with positive daily returns by investing in its shares . The calculation must be made for n = 0, 1, 2, 3, 4, 5.

b) Calculate the mean, variance and standard deviation of the exact number of days in a row with a positive return on investment in its shares.

c) Calculate the probability to achieve a positive return n or more days in a row by investing in the share. The calculation must be made for n= 0, 1, 2, 3, 4, 5.

d) Calculate, with previously acquired results from c) the probability for a positive return on any given workday in the investment of shares GIVEN/granted that the share has had positive returns the previous n days. The calculation must be made for n= 1, 2, 3, 4.
E(X)=P(X_i)*X_i
Var(X)=(X_i-E(X))²
I've been sitting with this assignment for hours, but just feel completely lost and have given up. To me it seems like they are, more or less, asking for the same thing in a) c) AND d), so yeah.. guess you can say probability isn't my strong side.

This is what I've done so far(though probably wrong):

P(0)=1-(P(1)+P(2)+P(3)+P(4)+P(5))
P(0)=1-(0,267+0,131+0,053+0,014+0,07)
P(0)=1-0,472
P(0)=0,528

n(0)=0,528
n(1)=1-P(0) =1-0,528=0,472
n(2)=1-P(0)+P(1) =1-0,795=0,205
n(3)=1-P(0)+P(1)+P(2) =1-0,926=0,074
n(4)=1-P(0)+P(1)+P(2)+P(3) =1-0,979=0,021
n(5)=1-P(0)+P(1)+P(2)+P(3)+P(4)=1-0,993=0,007As I said before, I have no idea if what I've done can be used or even what question I've answered (or have not answered)

b)
E(X)=0,528*0+0,267*1+0,131*2+0,053*3+0,014*4+0,007*5+0*6=0,779

Var(X)=0,528*(0-0,779)²+0,267*(1-0,779)²+0,131*(2-0,779)²+0,053*(3-0,779)²+0,014*(4-0,779)²+0,007*(5-0,779)²+0,0*(6-0,779)²=1,060159I'm not used to forums really, so I hope I've done this right. If not, just tell me, and I'll fix it.
Thanks in advance :)

 

[Ray Vickson]

I'm not a native English speaker, and this assignment wasn't originally in English, so I had to translate it to English, thus the grammar errors, but hopefully it's understandable)
-----------------------------------
For an investment in an equity assessed in general to be the following probabilities attached to the exact number of days in a row with positive returns

One day with a positive return , 2 day negative return 26.7%

2 consecutive days with positive returns , 3. day negative return……………. 13.1%

3 consecutive days with positive returns , 4. day negative return……………. 5.3%

4 consecutive days with positive returns , 5. day negative return …………….1.4%

5 consecutive days with positive returns , 6. day negative return …………….0.7%

6 or more consecutive days with positive returns …………………………………….0.0%

For example , there seem to be a probability of 26.7% on any given workday to achieve a positive return and then on the following day to achieve a negative return. Similarly, there seem to be a probability of 13.1% on any given workday to achieve a positive return and then on the following day also to achieve a positive return while on the third day to achieve a negative return.

a) Calculate by using the table above the probability of maximum n days in a row with positive daily returns by investing in its shares . The calculation must be made for n = 0, 1, 2, 3, 4, 5.

b) Calculate the mean, variance and standard deviation of the exact number of days in a row with a positive return on investment in its shares.

c) Calculate the probability to achieve a positive return n or more days in a row by investing in the share. The calculation must be made for n= 0, 1, 2, 3, 4, 5.

d) Calculate, with previously acquired results from c) the probability for a positive return on any given workday in the investment of shares GIVEN/granted that the share has had positive returns the previous n days. The calculation must be made for n= 1, 2, 3, 4.



E(X)=P(X_i)*X_i
Var(X)=(X_i-E(X))²



I've been sitting with this assignment for hours, but just feel completely lost and have given up. To me it seems like they are, more or less, asking for the same thing in a) c) AND d), so yeah.. guess you can say probability isn't my strong side.

This is what I've done so far(though probably wrong):

P(0)=1-(P(1)+P(2)+P(3)+P(4)+P(5))
P(0)=1-(0,267+0,131+0,053+0,014+0,07)
P(0)=1-0,472
P(0)=0,528

n(0)=0,528
n(1)=1-P(0) =1-0,528=0,472
n(2)=1-P(0)+P(1) =1-0,795=0,205
n(3)=1-P(0)+P(1)+P(2) =1-0,926=0,074
n(4)=1-P(0)+P(1)+P(2)+P(3) =1-0,979=0,021
n(5)=1-P(0)+P(1)+P(2)+P(3)+P(4)=1-0,993=0,007


As I said before, I have no idea if what I've done can be used or even what question I've answered (or have not answered)

b)
E(X)=0,528*0+0,267*1+0,131*2+0,053*3+0,014*4+0,007*5+0*6=0,779

Var(X)=0,528*(0-0,779)²+0,267*(1-0,779)²+0,131*(2-0,779)²+0,053*(3-0,779)²+0,014*(4-0,779)²+0,007*(5-0,779)²+0,0*(6-0,779)²=1,060159


I'm not used to forums really, so I hope I've done this right. If not, just tell me, and I'll fix it.
Thanks in advance :)

Are there only two "return" values (P and N)? That is, are all the positive returns the same, and all the negative returns the same? Furthermore, how is the multi-day return calculated: is it added (to get
\text{3-day return} = P_1 + P_2 - N_3
for example) or is it obtained through multiplication, as in
\text{3-day return} = (1+P_1/100)(1+P_2/100)(1-N_3/100)-1?
For small P_i/100 and N_i/100 there is not much difference between the two, but the second form corresponds more realistically to how it is done in financial institutions.

If all the P_i and N_i are different I don't think you have enough information to estimate the one-day return distribution with any kind of confidence.

Finally, the behavior of the returns argues against any type of simple probability model, because positive runs are consistently longer than negative runs in each P-N run pair, and that suggests that successive returns are far from random or independent. I suspect they are not even Markovian.

 

[anonymousk]

Thanks for the reply, but I'm still as lost as before.

 

[D H]

I've been sitting with this assignment for hours, but just feel completely lost and have given up. To me it seems like they are, more or less, asking for the same thing in a) c) AND d), so yeah.. guess you can say probability isn't my strong side.

These are very different questions.

(a) asks for the probability of N or fewer work days of consecutive gains, followed by a loss.
Hint: Note that the probabilities do not add to 100%. Why not?

(c) asks for the probability of N or more work days of consecutive gains (day N+1 may or may not be a loss).
Hint: The probability of 0 or more work days of consecutive gains is 100%. Why?

(d) asks for the probability of a gain on some work day given a gain on the N previous work days.
Hint: You'll need the formula for conditional probability to answer this question.

 

[anonymousk]

These are very different questions.

(a) asks for the probability of N or fewer work days of consecutive gains, followed by a loss.
Hint: Note that the probabilities do not add to 100%. Why not?

(c) asks for the probability of N or more work days of consecutive gains (day N+1 may or may not be a loss).
Hint: The probability of 0 or more work days of consecutive gains is 100%. Why?

(d) asks for the probability of a gain on some work day given a gain on the N previous work days.
Hint: You'll need the formula for conditional probability to answer this question.

(a): The probabilities do not add to 100% because we do not know P(0)? So I'm thinking I have to find P(0) by doing 1-(P(1)+P(2)+...+P(5)), but don't know where to go from there to find the probability for max/highest n days in a row with positive daily returns.

(c): Again, same with (a), if P(0) is included, the total probability is 100%?

(d): I have used the conditional probability formula before, but there seems to be more than 2 events here. I don't even know where to start putting in numbers.
EDIT: Okay, so I think I figured out (a)

a)

N(0)= 100-47,2=52,8 %
N(1)= 100-20,5=79,5 %
N(2)=100-7,4=92,6 %
N(3)= 100-2,1=97,9 %
N(4)=100-0,7=99,3 %
N(5)= 100-0=100 %

So now I've found via calculations that the number of max n days with a positive return is 5, correct? Now to c and d, hehe!

 

[D H]

Correct. Question (a) asked for P(N≤n). Question (c) is the other way around. It's asking for the probability of a positive return on n or more days. In other words, it's asking for P(N≥n). You already have the answer to question (c) in your answer to question (a).

 

[anonymousk]

Should it be N(0), N(1) like I wrote it before, or is the correct way to write it P(0) instead of N(0), etc? Just to be sure.

so it should be
P(0)= 100-47,2=52,8 %
P(1)= 100-20,5=79,5 %
P(2)=100-7,4=92,6 %
P(3)= 100-2,1=97,9 %
P(4)=100-0,7=99,3 %
P(5)= 100-0=100 %

And can you quickly clarify the difference in N and n? I'm having a hard time wrapping my head around this/probability.


Edit:

c)

p(0)=52,8
p(1)=100-P(0) =100-52,8=47,2
p(2)=1-P(0)+P(1) =100-79,5=20,5
p(3)=1-P(0)+P(1)+P(2) =100-92,6=7,4
p(4)=1-P(0)+P(1)+P(2)+P(3) =100-97,9=2,1
p(5)=1-P(0)+P(1)+P(2)+P(3)+P(4)=100-99,3=0,7

So this is the answer for c) ? (and again, it should be P(0) etc?

And thanks by the way! Your help is very much appreciated! :)

 

[D H]

And can you quickly clarify the difference in N and n? I'm having a hard time wrapping my head around this/probability.

That's a very standard notation. For example, P(X≤x) is a function of some variable x. What this shorthand means, in words, is "for a given value of x, what is the probability that a value drawn from the random variable X will have a value less than or equal to x."

You'll see P(X≤x) a lot. It comes up so often that it has been given a name. P(X≤x) is the cumulative probability function for the random variable X, or CDF for short. What question (a) was asking for was the CDF for this particular probability distribution.

Edit:

c)

p(0)=52,8
p(1)=100-P(0) =100-52,8=47,2
p(2)=1-P(0)+P(1) =100-79,5=20,5
p(3)=1-P(0)+P(1)+P(2) =100-92,6=7,4
p(4)=1-P(0)+P(1)+P(2)+P(3) =100-97,9=2,1
p(5)=1-P(0)+P(1)+P(2)+P(3)+P(4)=100-99,3=0,7

Close, but no cigar. Think about what the question is asking for.

Hint: You will *always* (i.e., probability = 100%) have 0 or more days without a loss.

 

[anonymousk]

so the correct way to write it, was it:
N(0)= 100-47,2=52,8 %
N(1)= 100-20,5=79,5 %
N(2)=100-7,4=92,6 %
N(3)= 100-2,1=97,9 %
N(4)=100-0,7=99,3 %
N(5)= 100-0=100 %

or with P(0)=100-47,2=52,8 % etc?
Just want to properly write it down.

giving c) another shot:

?(0)=100
?(1)=P(0)-P(1) =100-47,2=52,8
?(2)=P(0)-P(1)+P(2) =100-47,2-20,5=32,5
?(3)=P(0)-P(1)+P(2)+P(3) =100-47,2-20,5-7,4=24,9
?(4)=P(0)-P(1)+P(2)+P(3)+P(4) =100-47,2-20,5-7,4-2,1=22,8
?(5)=P(0)-P(1)+P(2)+P(3)+P(4)+P(5)=100-47,2-20,5-7,4-2,1-0,7=22,1

hmm, that's definitely wrong.EDIT: I gave it another shot, but if this is correct, shouldn't N(6), n(6) or P(6) (I really haven't gotten the hang of those yet) give 0? and not the 5th day like below?

?(0)=100-52,8=47,2
?(1)=100-79,5=20,5
?(2)=100-92,6=7,4
?(3)=100-97,9=2,1
?(4)=100-99,3=0,7
?(5)=100-100=0

(and again, i put the ? in there because I'm unsure of what letter to use. Is it the probability that i find now? Thus it should be P(X) ?)

 

[D H]

You were correct when you had a probability of zero or more days of consecutive gains as 100%. Think about it: How could this probability be anything *but* 100%? Zero or more days is the full sample space.

Look at n=1, the probability that you have one or more days of positive gains. One way to look at this is that you have exactly one day of positive gains (followed by a loss), or exactly two days of positive gains (followed by a loss), or ...

Another way to look at it is that you do *not* have n-1 or fewer days of consecutive gains. In other words, it's the opposite of question (a), but offset by one.

Yet another way to look at it is to build from large values of n on down. What's the probability you will have six or more days of consecutive gains? The probability you will have five or more days of consecutive gains is the probability you will have exactly five days of consecutive gains plus the probability you will have more than five days of consecutive gains. Note that "six or more days" is the same as "more than five days"; this is a discrete probability distribution. Now you should be able to answer the question for four or more days, then three or more, etc.

 

[anonymousk]

P(N>5)=0
P(N>4)=0+0,7=0,7
P(N>3)=0,7+1,4=2,1
P(N>2)=2,1+5,3=7,4
P(N>1)=7,4+13,1=20,5
P(N>0)=100

Very skeptical about this. We are currently 2 guys sitting here trying to figure this out. We feel like we're getting close, but it seems we need an extra hand. (not that we haven't already been given plenty of hands from you already ;P )EDIT:Another try:
P(N≥5)=0+0,7=0,7
P(N≥4)=0,7+0,7=1,4
P(N≥3)=2,1+1.4=3,5
P(N≥2)=7,4+3,5=10,9
P(N≥1)=20,5+10,9=31,4
P(N≥0)=100

 

[D H]

Neither one is correct.

Starting from n=6, P(N≥6)=0.The next step down is n=5. You can have a run of five straight days or longer with positive gains if you have a run of more than five straight days with positive gains (probability=0) or if you a run of five days with positive gains followed by a down day (probability = 0.7%). Thus P(N≥5)=0.7%.

A side question here: You can't always add probabilities. It's valid in this case. Why is that? What is the condition that let's me use P(N≥5) = P(N=5) + P(N>5)?

Another side question: I implicitly used P(N>5) = P(N≥6) in the above. What let me do that?The next step down is n=4. You can have a run of four straight days or longer with positive gains if you have a run of more than four straight days with positive gains (probability=0.7%) or if you have a run of four straight days with positive gains followed by a down day (probability = 1.4%). Thus P(N≥4)=2.1%.You should be able to take it from here.

 

[anonymousk]

ooooooh!

P(N≥5)=0.7%
P(N≥4)=0,7+1,4=2,1%
P(N≥3)=2,1+5,3=7,4%
P(N≥2)=7,4+13,1=20,5%
P(N≥1)=20,5+26,7=47,2
P(N≥0)=100%

This should be right now. Thanks a lot!

To your first side question: I'm guessing you can add the probabilities in this case because they are dependent on each other. As in, day 1 has to happen before day 2 can?

And second question: Reason you can set P(N>5) = P(N≥6) is we know that all days with a positive return above 5 have a 0 probability of happening, therefore you can set them equal. Or am I grasping at straws here?

Been looking at d) and the conditional probability formula, but have no idea what numbers/probabilities to use, since we have more than 2 events/days.

EDIT:

After taking a closer look at c), I managed to calculate the mean and variance (As in, I think I've used the correct numbers.)

E(X)=0,528*0+0,267*1+0,131*2+0,053*3+0,014*4+0,007*5+0*6=0,779

Var(X)=0,528*(0-0,779)^2+0,267*(1-0,779)^2+0,131*(2-0,779)^2+0,053*(3-0,779)^2+0,014*(4-0,779)^2+0,007*(5-0,779)^2+0,0*(6-0,779)^2=1,060159

And the standard deviation should be the squared root of the variance, the squared root of 1,060159.

However, I'm still stuck on d)

 

[D H]

Correct on (c), and my first side question. More generally, you can add probabilities of two events if and only if the two events are mutually exclusive. Another way to look at it is that the events represent disjoint, or non-intersecting, sets. Probability and statistics are difficult topics if you don't understand basic set theory. It helps a lot to know at least a bit about set theory.

So why can you equate P(N>5) with P(N≥6)? They're the same set!With regard to (d), whenever you see the word GIVEN you should think conditional probability or Bayes' theorem. You probably haven't been taught the latter yet, so you should think conditional probability. The probability of some event B given that some event A has occurred is the probability of the intersection of the two events divided by the probability of event A: $P(B|A) = \frac{P(A\cap B)}{P(A)}$.

Question (d) asks you to compute the probability of a positive return given that you had positive returns on the previous n days. You were also given a hint to use the results from question (c) to answer this question.

What are the events A and B that will help attack this problem, and what do they have to do with the results from question (c)?

 

[anonymousk]

aah.. I'm thinking P(A∩B) is the added values of the numbers I've been told to calculate from c), so:

47,2+20,5+7,4+2,1 = 77,2
P(A∩B∩C∩D) = P(n(1)∩n(2)∩n(3)∩n(4))=77,2.


$P(B|A) = \frac{77,2}{P(A)}$.



And P(A) is in this case the probability for the 5th day? Hmpf, I've probably gone completely off track again. Having a hard time understanding this question aswell.

 

[D H]

You are asked to solve that conditional probability for n=1, 2, 3, and 4. Those are four different probabilities. You'll have to compute each one separately.

Question #1: For some specific value of n, what does "GIVEN/granted that the share has had positive returns the previous n days" mean?

This is what I labeled as event "A" in my previous post. In particular, what does it mean with regard to your answer to question (c), and given that, what is the probability of event A? Note: This will depend on the value of n.Question #2: What is your event B?

You are looking to calculate the probability of at least one more day with a positive gain given that you have already had n successive days of positive gains. In particular, what does this mean with regard to your answer to question (c)?Question #3: What is the intersection between events A and B?

This is a bit of a trick question. The answer is easy once you think about it. A Venn diagram may be of help.Question #4: What is "the probability for a positive return on any given workday in the investment of shares GIVEN/granted that the share has had positive returns the previous n days"?

This should be easy if you have answered the above questions. Note that this is the question you are supposed to answer. Notice how it helps to break the question down into little parts.

 

[anonymousk]

I would normally call myself above average in math, but I'm just having a really hard time wrapping my head around this.

#1: I'm thinking event A in this case (when the value of n=1) is P(N=1)=26,7, then number we are given to start with.
#2: and event B is from the answer in c), P(N≥1)=47,2
#3: intersection would be the chances of both happening at the same time, right? hmm.

 

[D H]

I would normally call myself above average in math, but I'm just having a really hard time wrapping my head around this.

#1: I'm thinking event A in this case (when the value of n=1) is P(N=1)=26,7, then number we are given to start with.

That is absolutely the wrong event. The probability that the next day will realize a positive gain is identically zero given the event that you cited. Remember that those events (the ones with probabilities of 26.7%, 13.1%, 5.3%, etc.) were defined as being n successive days of positive gains followed by a day of negative returns.

#2: and event B is from the answer in c), P(N≥1)=47,2

This is not event B. This is event A for n=1. Event A for n=1 is the even that "the share has had positive returns the previous 1 days". It might continue to show positive returns. You, the investor, want it to continue to show positive returns. Event A for some value n is the random variable N having a value greater than or equal to n, or N≥n.

So given that this is event A, what is event B? Hint: The answer to this question also lies in the answer to question (c).

 

[anonymousk]

Event B is the probability for the following day ,P(N≥2)=20,5 then I believe.
Still lost on the intersection though.

EDIT: It's 4am here, and brain is barely functioning anymore, so if I stop replying, it's because I've gone to sleep, but thanks again for all your help (and patience).

2nd EDIT: Hmm wait, could the intersection be 47,2-20,5?

 

[D H]

Correct on event B.

How to visualize the intersection? On the one hand, event A represents one or more days of consecutive positive gains, and on the other, event B represents two or more days of consecutive positive gains. Think about the way you answered question (c). There's a definite relationship between this event A and event B. Draw a Venn diagram. They help some people a lot. Venn diagrams can be of considerable aid if you are one of those who think visually (right brain dominant).

 

[anonymousk]

Staring at a Venn diagram at the moment.. Could the intersection be 47,2? Grasping at straws again i think.. I'm going to sleep on this, and hope to solve this assignment tomorrow.

(again, thanks for the help)

 

[D H]

Sleep on it, and you are welcome.

Don't think of the intersection as a number, a probability. Think of it as exactly what the Venn diagram shows.

Try plotting the five mutually exclusive events outlined in the problem plus the one event that wasn't specified (loss on the very first day) as mutually exclusive events that encompass the sample space.

Note that I said five rather than six mutually exclusive events. The problem specified that the probability of "six or more consecutive days with positive returns" is zero. A zero probability event, at least at the introductory probability and statistics level, is a non-event. It cannot happen. That event is a non-event. It's not a part of the sample space. The mutually exclusive events are a loss on the very first day, one day of positive gains followed by a loss, two consecutive days of positive gains followed by a loss, (2, 3, 4 days of gains followed by a loss), and finally, five consecutive days of positive gains followed by a loss.You apparently purchased a Danish stock. (Denmark is lauded as the happiest and most egalitarian country in the world.) You should see positive gains more often than not; otherwise stockholders wouldn't be happy. Positive gains for two, maybe even three days in a row? That's okay, too. Four or five days in a row? That's getting to be a bit standoffish; the conditional probability drops. Positive gains for six days or longer? In Huxley's words, that's double-plus ungood.

 

[anonymousk]

Good morning (or good day here).

So I'm looking at the Venn diagram again. Since they are mutually exclusive, they have no intersection? They both can't happen at the same time. (or I might be doing something as simple as a Venn diagram wrong :P ).

Loss on the very first day, I'm not sure if it's the 52,8 ot 47,2, hm.

And hah, this is a Danish stock indeed :)

 

[anonymousk]

Just to understand this assignment better, can you tell me if the P(n)'s are correctly written?

P(N=n) is the probability to get exactly n amount of days where the following day gives a negative return.
P(N=0)=52,8% (not sure about this one)
P(N=1)=26,7%
P(N=2)=13,1%
P(N=3)=5,3%
P(N=4)=1,4%
P(N=5)=0,7%
P(N=6)=0%

P(N≤n):This is like the reverse probability or how you say it. The probability to get less than 3 days is 97,9%. This is the cumulative distribution function.
P(N≤0)=52,8%
P(N≤1)=79,5%
P(N≤2)=92,6%
P(N≤3)=97,9%
P(N≤4)=99,3%
P(N≤5)=100%

P(N≥n): having a bit hard time explaining this one. The probability to get 3 or more days is 7,4. Why is this number higher than probability to get exactly 3 days?
P(N≥5)=0.7%
P(N≥4)=0,7+1,4=2,1%
P(N≥3)=2,1+5,3=7,4%
P(N≥2)=7,4+13,1=20,5%
P(N≥1)=20,5+26,7=47,2
P(N≥0)=100%

Still kind of lost on the intersection. Even with a Venn diagram.

 

[HoldenJ]

I'm sitting with the same assignment and I have ended up with the same values as anonymousk.
But question b) also ask: According to Chebyshev's Theorem what does the calculated mean and standard deviation say about the exact amount of days in row with a positive return.
Thx in advance.

 

[D H]

HoldenJ, you need to show some work, just as anonymousk did. We do not do your homework for you at this site. We help you do your own homework.

It's not that we're lazy. It's exactly the opposite. We could have blurted out the answers to anonymousk's questions in post #2 and have been done with this thread. That would have been the easy way out for us. Doing that would not have helped anonymousk one single thing. He or she would have the answers to the homework, but that's it. When it comes time to sit an exam, or to use these concepts in a job, he or she would be clueless -- and we would not be there to help.

 

[HoldenJ]

Of course, I understand. My calculations are:
a)
P(N≤0)=100%-26,7%+13,1%+5,3%+1,4%+0,7%+0,0%=52,8%
P(N≤1)=100%-13,1%+5,3%+1,4%+0,7%+0,0%=79,5%
P(N≤2)=100%-5,3%+1,4%+0,7%+0,0%=92,6%
P(N≤3)=100%-1,4%+0,7%+0,0%=97,9%
P(N≤4)=100%-0,7%+0,0%=99,3%
P(N≤5)=100%-0,0%=100%

b)
Mean: E(X)=0,267*1+0,131*2+0,053*3+0,014*4+4*0,007+5*0,0=0,779
Variance: Var(X)=0,528*(0-0,779)2+0,267*(1-0,779)2+0,131*(2-0,779)2+0,053*(3-0,779)2+0,014*(4-0,779)2+0,007*(5-0,779)2+0,0*(6-0,779)2=1,060159
StdDev:1,060159^0,5=1,0296402

Then they ask according to Chebyshev's Theorem what does Mean an StdDev say about the exact amount of days in row with positive return.

I have benn struggeling with this for long time and it does not give any sense to me, since the StdDev is larger than the mean.

Srry again for the fast question without showing my calculations, but I am just desperate to get some hints that will send me in the right direction.

 

[D H]

Just to understand this assignment better, can you tell me if the P(n)'s are correctly written?

P(N=n) is the probability to get exactly n amount of days where the following day gives a negative return.
P(N=0)=52,8% (not sure about this one)
P(N=1)=26,7%
P(N=2)=13,1%
P(N=3)=5,3%
P(N=4)=1,4%
P(N=5)=0,7%
P(N=6)=0%

Almost. The last one is P(N≥6)=0%. Look at what you wrote in the original post. The last item is worded differently than all the others. It says that the probability of "6 or more consecutive days with positive returns" is 0%.

So how to infer that 52.8% for N=0? The question is about what happens today and in the future. The past? Yesterday was a losing day, but it's past. The goal is to see what's in store for us now and in the future. This means N is a non-negative number. We're also only concerned with whole days. This means N is a non-negative integer. Finally, since the probability for N≥6 is identically zero, the sample space is effectively the integers between zero and five, inclusive. Why inclusive? The given probabilities don't add to 1, and those events are mutually exclusive. They do not span the sample space. There's one missing piece of missing information, the probability that today will also be a losing day.

P(N≤n):This is like the reverse probability or how you say it. The probability to get less than 3 days is 97,9%. This is the cumulative distribution function.
P(N≤0)=52,8%
P(N≤1)=79,5%
P(N≤2)=92,6%
P(N≤3)=97,9%
P(N≤4)=99,3%
P(N≤5)=100%

Correct.

P(N≥n): having a bit hard time explaining this one. The probability to get 3 or more days is 7,4. Why is this number higher than probability to get exactly 3 days?

Because "three or more" means getting exactly three days, or exactly four days, or exactly five days of consecutive gains (followed by a loss).

P(N≥5)=0.7%
P(N≥4)=0,7+1,4=2,1%
P(N≥3)=2,1+5,3=7,4%
P(N≥2)=7,4+13,1=20,5%
P(N≥1)=20,5+26,7=47,2
P(N≥0)=100%

Correct.

Still kind of lost on the intersection. Even with a Venn diagram.

You haven't said what event B is yet. How can you calculate the intersection of some set with an unknown set?

 

[D H]

b)
Mean: E(X)=0,267*1+0,131*2+0,053*3+0,014*4+4*0,007+5*0,0=0,779

You have an error here. It happened where you switched from using probability*day number to using day number*probability. This mistake in the calculation of the mean also makes your standard deviation incorrect.

Then they ask according to Chebyshev's Theorem what does Mean an StdDev say about the exact amount of days in row with positive return.

What is Chebyshev's Theorem to you? (There are several theorems called Chebyshev's Theorem.) One of the theorems named after Chebyshev talks about bounds on the range of a random variable, even if you don't know the distribution. Is that what you are being asked to apply?

Also, how is the question worded? The phrases "exact amount" and "random variable" don't quite go hand in hand.

 

[anonymousk]

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.

 

[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.