Can anyone help to solve those questions?

[achiu]

I was told by my friend the following questions are from recent past exam (don't know which yr) but don't know whether this is true??? Anyway, I found they are interesting. Can anyone solve them? Thanks you!!

1. The number of severe storms that strike city J in a year follows a binomial distribution with n=5 and p=0.6. Given that m severe storms strike city J in a year, the number of severe storms that strike city K in the same year is m with probability 1/2, m+1 with probability 1/3, m+2 with probability 1/6.
Calculate the expected number of severe storms that strike city J in a year during which 5 severe storms strike city K.


2. The random variable Y1=e^X1 characterizes an insurer's annual property losses, where X1 is normally distributed with mean 16 and standard deviation 1.5. Similarly, the random variable Y2=eX2 characterizes the insurer's annual liability losses, where X2 is normally distributed with mean 15 and standard deviation 2.
The insurer's annual property losses are independent of its annual liability losses. Calculate the probability that, in a given year, the minimum of the insurer's property losses and liability losses exceeds e^16.


3. Let X be a continuous random variable with probability density function f(x)=te^(-tx) for x>0 and t>0. Let Y be the smallest interger greater than or equal to X. Determine the probability function of Y.

[Gandalf]

Originally Posted by achiu

1. The number of severe storms that strike city J in a year follows a binomial distribution with n=5 and p=0.6. Given that m severe storms strike city J in a year, the number of severe storms that strike city K in the same year is m with probability 1/2, m+1 with probability 1/3, m+2 with probability 1/6.
Calculate the expected number of severe storms that strike city J in a year during which 5 severe storms strike city K.

I think you need to derive the conditional distribution of J (severe storms hitting city J) given K (severe storms hitting city K) = 5.
Start by computing the unconditional joint densities for the case where k = 5:

f(3,5) = 1/6 p(j=3)
f(4,5) = 1/3 p(j=4)
f(5,5) = 1/2 p(j=5)

g(5) = marginal density for k=5 = sum of those

f(j|5) = f(j,5)/g(5), for j=3,4,5

E(J|5) = sum j=3 to 5 of j f(j|5)

2. The random variable Y1=e^X1 characterizes an insurer's annual property losses, where X1 is normally distributed with mean 16 and standard deviation 1.5. Similarly, the random variable Y2=eX2 characterizes the insurer's annual liability losses, where X2 is normally distributed with mean 15 and standard deviation 2.
The insurer's annual property losses are independent of its annual liability losses. Calculate the probability that, in a given year, the minimum of the insurer's property losses and liability losses exceeds e^16.

When is the minimum less than e^16? It happens unless Y1 and Y2 are each > e^16.
Pr(Y1>e^16) = Pr(X1>16)
Pr(Y2>e^16) = Pr(X2>16)
With a standard normal distribution table, you should be able to determine each of those.

Independent, so multiply to get the probabilty both are greater. Subtract result from 1 to get probability at least one is less (so that the minimum is less)

3. Let X be a continuous random variable with probability density function f(x)=te^(-tx) for x>0 and t>0. Let Y be the smallest interger greater than or equal to X. Determine the probability function of Y.

One way is to start by getting the cumulative distribution function of X, F(x). Then for any integer k, the cumulative probability function for k is
P(k) = F(k) (P(k) exists only for integers).
Then the probability function is P(k) - P(k-1).

Alternatively, what values of X correspond to K=k? Ans: k-1 < x <= k
What is the probability X is in that interval: integral from k-1 to k of te^(-tx) dx.

[achiu]

Thank you so much!

[Generalshamu]

Originally Posted by Gandalf

When is the minimum less than e^16? It happens unless Y1 and Y2 are each > e^16.
Pr(Y1>e^16) = Pr(X1>16)
Pr(Y2>e^16) = Pr(X2>16)
With a standard normal distribution table, you should be able to determine each of those.

Independent, so multiply to get the probabilty both are greater. Subtract result from 1 to get probability at least one is less (so that the minimum is less)

Wait don't you have to do X = ln Y here, since it follows a lognormal distribution???
From that you get Phi((ln x - mean )/standard deviation)?
For my answer I got .15425???

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Just for clarification, is the answer to the first question, 3.897637795?

[Colymbosathon ecplecticos]

Originally Posted by Generalshamu

Wait don't you have to do X = ln Y here, since it follows a lognormal distribution???
From that you get Phi((ln x - mean )/standard deviation)?
For my answer I got .15425???

-----

Just for clarification, is the answer to the first question, 3.897637795?

Recall: Y1 = exp(X1). When is Y1>exp(16)?

Right, when X1 > 16, because exp is monotone increasing.

[Generalshamu]

Oh but you get the same Z-scores of you do it this way...is it wrong to take it as a lognormal distribution?

[atomic]

Originally Posted by Generalshamu

Oh but you get the same Z-scores of you do it this way...is it wrong to take it as a lognormal distribution?

Because the logarithmic and exponential functions are order-preserving, one-to-one mappings, it's not necessary to deal with the lognormal distribution of Y. It suffices to simply look at the normal distribution of X. That is to say, the minimum of Y1 and Y2 is the same as the exponential of the minimum of X1 and X2, or

.

[Generalshamu]

Can you do that with ANY density function?

Where if they both have the same factors, you can just reduce the densities accordingly?

[atomic]

Originally Posted by Generalshamu

Can you do that with ANY density function?

Where if they both have the same factors, you can just reduce the densities accordingly?

I'm not sure what you mean by "reduce the densities accordingly."

If your question is whether the content of my previous post depends on the density of X1 and X2, the answer is no; all that is required is that the same order-preserving, one-to-one mapping be applied to each distribution. The key relationship is that g(a) < g(b) if and only if a < b, for a, b in the support of X1, X2, ..., Xn. Then min{g(X1), g(X2), ..., g(Xn)} = g(min{X1, X2, ..., Xn}).

If I may ask, is your background in mathematics? What I find curious is that you seem to have a propensity for asking questions that imply you are wanting to find very general yet powerful rules for working with probability distributions--things that you can apply in every case, while avoiding dealing with first principles. In my opinion, this is an unwise approach to understanding mathematics. Rather, a broad, solid foundation in the basics of mathematical reasoning is what allows one to develop "shortcuts" while still being aware of when they can be applied, or how they arise. I realize that this sort of learning is not the most time-efficient approach for the exams, but it is again in my view the best approach over the long run.

[Gandalf]

This is sort of saying what atomic did, and relating to my earlier post about looking at x instead of y. There are lots of situations where you have a condition like g(x) > k, or k1 < g(x) < k2, etc.

In those situations, it's often a good idea to try to figure out which values of x would make the inequality involving g(x) true. It tends to be easiest to do that if g(x) is a 1-to-1 increasing function, but the idea of figuring out what values of x work can be helpful in other situations, too.

E.g., if you want to know where g(x) = x^2 > 4, it is true when x < -2 or x > 2. Often that is a productive way to attack the problem.