Actuarial Outpost ASM 1st Edition (Exam P, Weishaus) Practice Exam 2 Q14, 16,24
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#11
12-28-2017, 03:49 PM
 dslee85 SOA Join Date: Apr 2017 College: Carnegi Posts: 14

Quote:
 Originally Posted by Z3ta That's correct. $U \in (3,\frac{10}{3})$ reads "U is in the interval between 3 and 10/3". It is just another way of writing $3 < U < \frac{10}{3}$ I worked backwards to determine the numbers 10/3 and 5. I knew that there would be certainty of $X\geq10$ if $2U\geq10 \rightarrow U \geq 5$ since 2U is its lower bound. Similarly, there is certainty that $X\not\geq10$ if $3U\leq10 \rightarrow U\leq\frac{10}{3}$ since 3U is its upper bound. The problem becomes a weighted sum of the conditional probabilities in each of those cases (weighted by the unconditional probability of each case) using the law of total probability. That's the discrete version. The integral itself is an application of the continuous version of the law of total probability (you integrate $P(X\ge10\mid U)$ times the PDF of $U$ over the possible values of $U$). Note: here $\color{red}U$ is really $\color{red}U \mid U \in (\frac{10}{3},5)$ since we're within a specific case of our first use of the law of total probability... but I thought it might be confusing if I wrote that. Does that make sense?
So if I'm understanding this correctly this is what you are saying.

I have to break up the range into 3 parts and then find the probability of each section and multiply by the conditional density.

Therefore, the answer should look something like this.

P(3<U<10/3|X>10)*P(X>10) + P(10/3<U<5|X>10)*P(X>10) + P(5<U<6|X>10)*P(X>10) =

Thank you again for your input and wish you happy holidays and a happy new year!
#12
12-28-2017, 04:27 PM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

Quote:
 Originally Posted by dslee85 So if I'm understanding this correctly this is what you are saying. I have to break up the range into 3 parts and then find the probability of each section and multiply by the conditional density. Therefore, the answer should look something like this. P(310)*P(X>10) + P(10/310)*P(X>10) + P(510)*P(X>10) = Thank you again for your input and wish you happy holidays and a happy new year!
You've got it a little mixed up dslee85.

The overall sum is in this form:
$P(X\geq10\mid 3

Here is the sum in the same order:
$P(X\geq10)=0\cdot\frac{1}{9}+\int_{10/3}^{5} \frac{3x-10}{3x-2x}\cdot\frac{1}{5-\frac{10}{3}} \, dx \cdot \frac{5}{9}+ 1\cdot\frac{1}{3}$

Then the integral within is an application of the continuous law of total probability.
$P(A)=\int P(A\mid u)f(u) \, du$

Here A is the event X>=10 given 10/3<U<5 and the pdf is of U|(10/3<U<5)... it's just more complicated looking because we're within an outer use of the law of total probability.

Happy Holidays and Happy New Year to you as well!

Last edited by Z3ta; 12-28-2017 at 08:11 PM..
#13
12-28-2017, 06:02 PM
 dslee85 SOA Join Date: Apr 2017 College: Carnegi Posts: 14

Quote:
 Originally Posted by Z3ta You've got it a little mixed up dslee85. The overall sum is in this form: $P(X\geq10\mid 3 My sum has each of those products in the reverse order: Then the integral within is an application of the continuous law of total probability. $P(A)=\int P(A\mid u)f(u) \, du$ Here A is the event X>=10 given 10/3
The solution you posted is not showing. Below "My sum has each of those products in the reverse order:" Could you upload it differently?
Thank you.
#14
12-28-2017, 06:21 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 30,855

Quote:
 Originally Posted by dslee85 So in summary, I should have 3 integrations where one ranges from 3 ~ 10/3, and another one between 10/3 ~ 5, and the last one between 5 ~ 6. Got it! Thank you very much for your explanation! Happy holidays and a happy new year!
Close to what I said. Even closer is that you need to think about the integrals for each of those three ranges (of the outer integral of the double integrals). Determining the upper and lower limits for each of those inside integrals will help you. Happy holidays and a happy new year to you, too, and good luck on the exam.
#15
12-28-2017, 08:07 PM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

Quote:
 Originally Posted by dslee85 The solution you posted is not showing. Below "My sum has each of those products in the reverse order:" Could you upload it differently? Thank you.
I edited my post. Can you see it now? I just rewrote part of my solution for reference there.