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P1.T2.300.1 Probability functions Question

PortoMarco79

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Thread starter #1
Hi

It has been 15 years since I last touched this stuff. I am very much not up to snuff any more.

Can someone assist?

How does: f(x) = x/8 - 0.75

become: F(X) = x^2/16 - 0.75*x + c.

Thanks
 

brian.field

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#2
f(x) is a probability density function. The integral of pdf from -infinity to +infinity must equal 1.

Capital F represents the Cumulative Density Function of CDF which is also represented by the integral of the pdf from -infinity to a and is interpreted as the probability that X <= a.

As a convention, f(x) represents the pdf and the corresponding capital letter, F(x) in this case, represents the CDF.
 

ShaktiRathore

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#3
Hi,
Its the simple integration of the f(x) , probability density function which should yield Cumulative Density Function F(x),
F(x)=CDF= integration of f(x)= x/8 - 0.75
F(x) = integration of[(x/8)dx - 0.75 dx]
F(x) = integration of[(xdx/8) - 0.75 dx]
F(x) = ((x^2/2)/8) - 0.75x +c sin ce the integration of x is x^2/2.
F(x) = (x^2/16) - 0.75x +c
THanks
 

ravinaghotra

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#4
Hello, can you please help and let me if we can use calculator so solve the above. I am using BA II Plus. What keys should be entered on the calculator.
 

brian.field

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#5
This is not really a calculator type problem - you need to know how to integrate the function which is something that you need to be able to using calculus. (Although you could probably pass this exam without any calculus, in other words, I wouldn't spend time reviewing calculus instead of other FRM aims.)
 

ShaktiRathore

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#6
Hi,
Yes I agree with Brian this types of problem shall not come where integration of some function is being asked only basic math shall come ,it is possible that some exceptional problem may come around so its better that you take some fresher course in calculus,as many concepts also demand understanding of basic fresher man calculus. Calculator BA II Plus has not integrate function i think.
also visit to see David views: https://www.bionicturtle.com/forum/...entiate-integrate-equations-in-the-exam.8814/
thanks
 

desh

New Member
#7
A Simple question is disturbing me.. I don't know why but need more clarifications....

Q: upload_2016-8-17_22-45-14.png
is :
The correct answer is : 0.80
Please explain in details so that I can get it easily... @brian.field Please help
 

PortoMarco79

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Thread starter #10
This is really not my strong point but, I figure that since you have 6 chances out of 10 to pick an industrial stock and 8 chances out of 10 to pick an industrial bond, you have 14 chances out of 20 to pick an industrial security.

14/20 = 0.7
 

PortoMarco79

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Thread starter #12
Ok. I misread the question. Sorry.

You are trying to pick a Bond (B) and an Industrial security (I).

P(B) = 10/20 (10 bonds in the pool 10 equities in the pool)
P (I) = 14/20 (6 utilities in the pool, 14 industrials in the pool)

So, you are trying to figure out the P (B & I) --which is the probability of picking B (10) + the probability of picking I (14) minus the probability of picking B or I (avoid double counting) (8)

10 + 14 - 8 = 16

16 / 20 = 0.8

I think this image sums it up best:

 

PortoMarco79

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Thread starter #13
Can some of the more polished forum members please validate my statement above? I am really not confident in my probabilities!
 

ami44

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#14
Can some of the more polished forum members please validate my statement above? I am really not confident in my probabilities!
PortoMarco79,

your answer is absolutly correct and is the intended answer.

Maybe a little easier way to come to the result is to count all the securities that are bonds or issued by industral firms, which are 16 (10 bonds + 6 stocks) and 16/20 = 0.8

But your way Matches the original answer.

Minor Nitpick, you write:
minus the probability of picking B or I
Imho this should be
"minus the probability of picking B and I"
 
#16
Hi PortoMarco, can you please put in the numbers in the image please..at least a description. I ma not able to figure out the probability of picking B&I in the venn image
 
#17
Assume a loss severity given by (x) can be characterized by a probability density function (pdf) on the domain [1, e^5]. For example, the minimum loss severity = $1 and the maximum possible loss severity = exp (5)~=$148.41. The pdf is given by f(x) = c/x as follows: f (x) = C/X s.t. 1 < x < e^5 where x =|loss severity |

What is the 95% value at risk (Var) i.e. given that losses are expected in positive values, at what loss severity value (x) is only 5 percent of the distribution greater than (x).

Please elaborate and explain the above question and provide easy solution @brian.field
 

ShaktiRathore

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#18
Hi,
Let 95% value at risk (Var) be X.
integration from 1 to e^5( f(x)dx) =1=> integration from 1 to e^5 (c ln(x) )=> c(5-0)=1=>c=1/5
integration from X to e^5 (f(x)dx) = integration from X to e^5 (c/x dx) = integration from X to e^5 (c ln(x)) = (1/5)*(-ln(X)+5)=.05 =>lnX=4.75 =>X=e^4.75 = $ 115.58
Thus at $ 115.58 only 5 percent of the distribution greater than (x).

thanks
 
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