Suppose that in `good years\', the number of claims fi led in the office of an

Question

Suppose that in `good years', the number of claims filed in the office of an insurance company arrive according to a Poisson process at the rate of 1 per day, while in `bad years', they arrive according to a Poisson process at the rate of 4 per day. Assume that next year will be a `good year' with probability 0.25. Let N(t) denote the number of claims filed during the first t days of next year. (a) (5 points) Find an expression for P[N(t) = n]. (b) (5 points) Is N(t) = n a Poisson process? Explain. (c) (5 points) Does N(t) = n have stationary increments? Why or why not? (d) (5 points) Does N(t) = n have independent increments? Why or why not? (e) (5 points) If next year starts off with exactly one claim by day 1, what is the conditional probability that it is a good year?

Explanation / Answer

THIS WILL BE HELPFUL FOR YOU Life is full of random events! You need to get a "feel" for them to be a smart and successful person. Independent Events Events can be "Independent", meaning each event is not affected by any other events. Example: Tossing a coin. Each toss of a coin is a perfect isolated thing. What it did in the past will not affect the current toss. The chance is simply 1-in-2, or 50%, just like ANY toss of the coin. So each toss is an Independent Event. Dependent Events But events can also be "dependent" ... which means they can be affected by previous events ... Example: Marbles in a Bag 2 blue and 3 red marbles are in a bag. What are the chances of getting a blue marble? The chance is 2 in 5 But after taking one out you change the chances! So the next time: if you got a red marble before, then the chance of a blue marble next is 2 in 4 if you got a blue marble before, then the chance of a blue marble next is 1 in 4 See how the chances change each time? Each event depends on what happened in the previous event, and is called dependent. That is the kind of thing we will be looking at here. "Replacement" Note: if you had replaced the marbles in the bag each time, then the chances would not have changed and the events would be independent: With Replacement: the events are Independent (the chances don't change) Without Replacement: the events are Dependent (the chances change) Tree Diagram A Tree Diagram: is a wonderful way to picture what is going on, so let's build one for our marbles example. There is a 2/5 chance of pulling out a Blue marble, and a 3/5 chance for Red: We can even go one step further and see what happens when we select a second marble: If a blue marble was selected first there is now a 1/4 chance of getting a blue marble and a 3/4 chance of getting a red marble. If a red marble was selected first there is now a 2/4 chance of getting a blue marble and a 2/4 chance of getting a red marble. Now we can answer questions like "What are the chances of drawing 2 blue marbles?" Answer: it is a 2/5 chance followed by a 1/4 chance: Did you see how we multiplied the chances? And got 1/10 as a result. The chances of drawing 2 blue marbles is 1/10 Notation We love notation in mathematics! It means we can then use the power of algebra to play around with the ideas. So here is the notation for probability: P(A) means "Probability Of Event A" In our marbles example Event A is "get a Blue Marble first" with a probability of 2/5: P(A) = 2/5 And Event B is "get a Blue Marble second" ... but for that we have 2 choices: If we got a Blue Marble first the chance is now 1/4 If we got a Red Marble first the chance is now 2/4 So we have to say which one we want, and use the symbol "|" to mean "given": P(B|A) means "Event B given Event A" In other words, event A has already happened, now what is the chance of event B? P(B|A) is also called the "Conditional Probability" of B given A. And in our case: P(B|A) = 1/4 So the probability of getting 2 blue marbles is: And we write it as "Probability of event A and event B equals the probability of event A times the probability of event B given event A" Let's do the next example using only notation: Example: Drawing 2 Kings from a Deck Event A is drawing a King first, and Event B is drawing a King second. For the first card the chance of drawing a King is 4 out of 52 P(A) = 4/52 But after removing a King from the deck the probability of the 2nd card drawn is less likely to be a King (only 3 of the 51 cards left are Kings): P(B|A) = 3/51 And so: P(A and B) = P(A) x P(B|A) = (4/52) x (3/51) = 12/2652 = 1/221 So the chance of getting 2 Kings is 1 in 221, or about 0.5% Finding Hidden Data Using Algebra we can also "change the subject" of the formula, like this: Start with: P(A and B) = P(A) x P(B|A) Swap sides: P(A) x P(B|A) = P(A and B) Divide by P(A): P(B|A) = P(A and B) / P(A) And we have another useful formula: "The probability of event B given event A equals the probability of event A and event B divided by the probability of event A Example: Ice Cream 70% of your friends like Chocolate, and 35% like Chocolate AND like Strawberry. What percent of those who like Chocolate also like Strawberry? P(Strawberry|Chocolate) = P(Chocolate and Strawberry) / P(Chocolate) 0.35 / 0.7 = 50% 50% of your friends who like Chocolate also like Strawberry Big Example: Soccer Game You are off to soccer, and want to be the Goalkeeper, but that depends who is the Coach today: with Coach Sam the probability of being Goalkeeper is 0.5 with Coach Alex the probability of being Goalkeeper is 0.3 Sam is Coach more often ... about 6 out of every 10 games (a probability of 0.6). So, what is the probability you will be a Goalkeeper today? Let's build a tree diagram. First we show the two possible coaches: Sam or Alex:

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