ACTL1101 Questions Help (mostly first year uni probability) (2 Viewers)

[InteGrand]

The above shows that memorylessness becomes equivalent to the condition that S(t+s) = S(t)S(s) for all t, s (where S is the survival function of the distribution). So to show that the only continuous distribution that is memoryless is Exponential, it suffices to show that continuous solutions to the above functional equation are only those that are of the form S(t) = e^{-lambda*t}. You can either try this as an exercise, or view the proof here: https://en.wikipedia.org/wiki/Memor...s_distribution_is_an_exponential_distribution .

 

[Rhinoz8142]

Something's weird...

Just by inspection I could tell that the expectation for the second case would be higher than the first. But why should I not be surprised that the variances are the exact same?

I have a feeling this isn't just by coincidence...

this is something similar to my finance subject, where we discuss risk and return.

 

[He-Mann]

Something's weird...

Just by inspection I could tell that the expectation for the second case would be higher than the first. But why should I not be surprised that the variances are the exact same?

I have a feeling this isn't just by coincidence...

Note that A = 25000 – B, where A is the r.v. describing the value from investment A, and similarly for B.

Therefore, by properties of variance (namely that an additive constant doesn't change the variance, and neither does the presence of a multiplicative factor of -1), we have

Var(A) = Var(25000 – B) = Var(B).

Referring to the above exchange, what is an intuitive interpretation of this variance property? Why is Var(k - X) = Var(X)? I understand the algebraic explanation but it gives me no further understanding of how variance works.

 

[InteGrand]

Referring to the above exchange, what is an intuitive interpretation of this variance property? Why is Var(k - X) = Var(X)? I understand the algebraic explanation but it gives me no further understanding of how variance works.

Intuitively speaking, Variance measures how much fluctuation there is in the values taken on by a random variable. A r.v. with very low variance will generally tend to take on values clustered close to the mean, whereas a r.v. with very high variance will tend to take on more "erratic" values about its mean.

If we firstly consider X and Y := X + c, where c is a constant, these two have same variance, and the intuition for this is that with Y, we are just shifting all the values X took up by c. This doesn't change the "erratic-ness" of the values taken, so the fluctuation about the mean is still the same, so the variance is the same.

By an essentially similar reason, Z := -X has the same variance as X. Intuitively, this is because all Z's values are the values taken by X but just "flipped" (negated). This means the overall (squared) fluctuations of the values of Z about its mean are still the same, so the variance is still the same.

So we know that adding constants or negating a r.v. preserve the variance. Combining these facts means that k - X and X will have the same variance.

 

[He-Mann]

Intuitively speaking, Variance measures how much fluctuation there is in the values taken on by a random variable. A r.v. with very low variance will generally tend to take on values clustered close to the mean, whereas a r.v. with very high variance will tend to take on more "erratic" values about its mean.

If we firstly consider X and Y := X + c, where c is a constant, these two have same variance, and the intuition for this is that with Y, we are just shifting all the values X took up by c. This doesn't change the "erratic-ness" of the values taken, so the fluctuation about the mean is still the same, so the variance is the same.

By a similar reason, Z := -X has the variance as X. Intuitively, this is because all Z's values are the values taken by X but just "flipped" (negated). This means the overall (squared) fluctuations of the values of Z about its mean are still the same, so the variance is still the same.

So we know that adding constants or negative a r.v. preserve the variance. Combining these facts means that k - X and X will have the same variance.

Shiet. That was beautiful. Thanks heaps for the explanation. Makes sense and brought more clarity to this topic.

 

[leehuan]

Answer's D

 

[InteGrand]

Answer's D

It suffices to find the expected payment in refunds for 1 printer and just multiply this by 100 (due to linearity of expectation)

To find this for 1 printer, calculate the following probabilities:

a := Probablity of failure in the first year (i.e. Pr(X < 1))
b := Probability of failure in second year (i.e. Pr(1 < X < 2)) (where X has the given density function).

The expected refund paid for one printer is then 200a + 100b.

 

[leehuan]

I'm pretty sure this is still under the wing of probability and statistics...

Anyway, I could handle the simple algebra they omitted to finish off the question. I'm just not sure as to how they generated that first expression; for some reason it did not follow for me despite the hint they gave.

Guidance please? I don't think it's hard, just don't see it.

 

[VBN2470]

Let p be the premium the individual is willing to pay so that he is indifferent between paying for the insurance or losing a random amount X following some distribution. We want find such a p such that E[u(w-p)] = E[u(w-X)]. Since the LHS expression is simply the expectation of a deterministic number, take it out of the expectation and re-arrange the equality to get the desired value of p (as shown in the solution).

 

[leehuan]

How does this solution work? I remember reading about it somewhere but I don't recall where and it just appears as magic..

 

[InteGrand]

How does this solution work? I remember reading about it somewhere but I don't recall where and it just appears as magic..

They basically used a probability generating function (PGF) (except not normalised, but that doesn't matter much, just divide through by a 6 for each PGF if you want, which is what they do in the end to get the probability). Recall that the PGF of a sum of independent random variables is the product of the individual PGF's. What they did is have coefficients as 'ways' rather than probabilities, so with their thing, just need to normalise it in the end if you want the probability.

 

[InteGrand]

You can read up more about PGF's and their properties here: https://en.wikipedia.org/wiki/Probability-generating_function .

 

[He-Mann]

This is only a guess as I've never seen this perspective on probability before. Please take this with a lot of skepticism.

Look to the sample space and let's do the first die. Sample space is 6 with 3 going to face 1, 2 going to face 2, and 1 going to face 3. Now, the number of events the first die can produce is:

(face1) or (face1) or (face1) or (face2) or (face2) or (face3) = 3*(face1) + 2*(face2) + (face3) = 3x^1 + 2x^2 + x^3 (call this event A).

where x^i represents the event where the i-th face pops up.

Similarly, for the second die, we get number of events: x^1 + 2x^2 + 3x^3 (call this event B).

We need, (event A) and (event B) which finds the total number of events that can happen when both die are thrown. It has 6*6 = 36 sample space.

So, (3x^1 + 2x^2 + x^3)(x^1 + 2x^2 + 3x^3) = ... = 3x^2 + 8x^3 + 14x^4 + 8x^5 + 3x^6. This can interpreted as,

3*(roll a total of 2) + 8*(roll a total of 3) + 14*(roll a total of 4) + ...

Clearly, rolling a total of 4 occupies most of the sample space relative to the other events hence that's the answer.

Prob = occurence of event / samples pace = 14/36.

 

[He-Mann]

They basically used a probability generating function (PGF) (except not normalised, but that doesn't matter much, just divide through by a 6 for each PGF if you want, which is what they do in the end to get the probability). Recall that the PGF of a sum of independent random variables is the product of the individual PGF's. What they did is have coefficients as 'ways' rather than probabilities, so with their thing, just need to normalise it in the end if you want the probability.

Interesting concept. I'm in second year,doing stats and never seen this but first year actuary do this.. shit.

Would you mind reviewing my interpretation?

 

[InteGrand]

Interesting concept. I'm in second year,doing stats and never seen this but first year actuary do this.. shit.

Would you mind reviewing my interpretation?

Basically what we do when we expand polynomial products (or products of power series) like this, to get the coefficient of x^{k}, we'll combine terms that'll give an exponent of k, which becomes like adding, due to index laws.

E.g. To get the coefficient of x^{5} in ((1/6)+(2/6)x+(3/6)x^{2})((3/4)x^{3} + (1/4)x^{4}), it's ((2/6)*(1/4) + (3/6)*(3/4)) = 11/24 (because we can combine the x in the first brackets with the x^{4} in the second, or the x^{2} from the first brackets with the x^{3} in the second. (So we're doing addition: combining all powers m and n from the two brackets such that m+n = 5). So if the coefficients here were probabilities, then we're adding up the probability all the ways we could get a total of 5 (if the first brackets represents the probability distribution of a die with faces 0, 1 and 2, with probability of showing 0, 1 and 2 being 1/6, 2/6 and 3/6 respectively, and similarly for the second brackets. In other words the coefficient of x^{5} in the expansion of the product gives us the probability of getting a total of 5.

In general, multiplying PGFs like this would give us probabilities for the sum of r.v.'s.

(This is essentially something known as convolution, by the way.)

 

[leehuan]

A coin with probability p of coming up heads is tossed until the rth head appears. The probability that the number of tosses required is n (n>=r) is given by...


This was a multiple choice question, but technically if I saw it I can just, without much further thought, think oh duh negative binomial right?

 

[InteGrand]

Yeah it's negative binomial.