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#101 Help Me ! » Parameter Estimation: Statistics » 2011-06-01 13:44:48

lindah
Replies: 23

http://img202.imageshack.us/i/64221633.png/

I have a feeling what I have done is wrong since one parameter shouldn't be expressed in another?

Thanks in advance for any feedback
Lin

#102 Re: Help Me ! » Joint distribution E[P] » 2011-04-26 00:01:54

lindah

I have figured this one out: it concerns properties of a bivariate normal density.

Thanks and Regards,
Linda

#103 Help Me ! » Joint distribution E[P] » 2011-04-24 19:13:19

lindah
Replies: 2

Hi guys,

Return of two stocks are joint normally with
mean of stock 1=5% , mean of stock 2 =4.4%, variance stock 1 =0.25, Variance stock 2 = 0.16
The correlation is 0.3.
The price of the first stock is $10.50 and the second is $15.
If invested in 2 stocks of firm 1 and 5 stocks of firm 2 - find the expected return.
If it is known that the return of stock 1 is 6%, find the return and standard deviation of stock 2.

I have calculated the expected return of the portfolio as 4.57%.

However I cannot figure out the second part of the question: finding the return and stdev of stock 2. I was thinking along the lines of using correlation/covariance formulas and solve for the missing variable, but seem to unable to do this as it all involves expected returns.

Thank you in advance for any feedback
Linda

#104 Re: Help Me ! » Transformation of Uniform random variable » 2011-04-23 13:03:30

lindah

Hi guys,

I happens that I found an example that confirms the latter approach outlined in my post above.
Apologies for cluttering up some space!!!

Linda

#105 Help Me ! » Transformation of Uniform random variable » 2011-04-23 12:49:07

lindah
Replies: 1

Hi guys,

A r.v. is uniform dist on [-1,1], The transformation is

, so find the density functions, and its first E[X] and second moments (var).

I have already found the cdf

, and pdf as
for interval
.

I'm quite sure this is a silly question faint but I wanted to confirm, since the examples I have seen so far are all linear transformations e.g. y=x/2
To find the expected value of Y, would this formula still apply?:


In this case 0.5?

Or because the pdf is no no longer linear, I would find the EV as (I get a different answer):

Thanks in advance for any feedback.
Lin

#106 Re: Help Me ! » Probability of default on loans » 2011-04-23 12:25:37

lindah

Hi bobbym;

Thanks for setting it out that way, it makes it so much clearer than (0.0082 x 0.9918)!!!
As always I try to over-complicate probability questions

Thank you very much.
Lin

#107 Help Me ! » Probability of default on loans » 2011-04-22 18:45:10

lindah
Replies: 3

Hi guys,

May I confirm if my working is correct for this question:

In a portfolio of loans 70% are rated ‘good’ with the probability of default of 0.001, 25% ‘standard’ with the probability of default 0.01 and 5% ‘stressed’ with the probability of default 0.1. If two loans are chosen at random what is the probability that exactly one of them will default?

I've calculated the probability of default as:
    P(default) = P(default | Good) P(Good)    + P(default | Standard) P(Standard)+ P(default |Stressed) P(Stressed)       
                   = (0.001 x 0.7) + (0.01 x 0.25) + (0.1 x 0.05)               
                   = 0.0082 or 41/5000
The probability of not defaulting is, per the complementary rule is:
        1 - 0.0082    = 0.9918

Now this is where I am not sure of myself:
Is the P(exactly one default) simply:
1) 0.0082 x 0.9918 or is it
1) 41/5000 x 4959/4999 - because of no replacement?

Thanks in advance for any feedback
Lin

#108 Re: Help Me ! » Random Variables question » 2010-08-29 09:00:04

lindah

Hi Bob and bobby,

I see I've overlooked repeating combinations.

Thank you very much for the clear explanations!!

Regards
Linda

#109 Help Me ! » Random Variables question » 2010-08-26 05:40:45

lindah
Replies: 5

Hi,

I am experiencing a brain freeze on how to approach the following question:

A group of five components contains two defective components. If components are taken from the five at a time and tested, let X be the trial at which the second defective is found.

a) What is the probability distribution of X?

b) Find the mean of X

My initial approach was to list the possible ordered scenarios {DDNNN, NNNDD, NDNDN} which would have 5! possible sequences, but then I am not sure how to derive a probability distribution.

Can someone point me in the right direction?

Thank you in advance for your help

#110 Help Me ! » Probability - different time frames » 2010-02-10 12:44:49

lindah
Replies: 2

Hi all,
May I verify if my thinking is spot on [I've exhausted the ideas, and this is the best I can come up with yikes]?

Thank you in advance for any assistance!

If the chances of seeing a meteor outside within an hour are 84%, what is the chance of seeing at least one new meteor outside in 30 minutes?

P(see meteor at least once within hr) = 1 - [P(seeing none in 30 min)]^2


= 0.6

and

If the probability of observing a car in 20 minutes is 609/625, What is the probability of observing at least one car in five minutes?

P(at least one car in 20 minutes) = 1 - [P(no cars in 5 minutes)]^4


#111 Re: Help Me ! » Expected value of a dice game » 2010-02-10 12:42:48

lindah

Thank you for your prompt responses!!

@mathsyperson
I think the question was geared towards the second interpretation

But thank you for the first scenario as well...it gives me more food for thought smile

#112 Help Me ! » Expected value of a dice game » 2010-02-09 16:18:13

lindah
Replies: 4

You can roll 2 die a maximum of 2 times. You will be paid the maximum value that you obtain either on the last roll  (either first or second roll).
What is the expected value?

May I ask if this is this simply (6+5+4+3+2+1)/6 = $3.50
Or will it involved conditional probabilities

E.g. if the first roll is {1,2,3} then I will most likely roll again.
if it is a {4,5,6} then I will take my money because it is greater than the expected value of $3.50

EDIT: in hindsight this may be the case - expected value of 4.25???
(0.5 x 3.5) + (0.5 x 5) = 4.25

#113 Re: Help Me ! » Counting unordered items » 2010-02-09 16:08:42

lindah

Thanks FL

Yes I keep coming up with the same conclusion as you 280
What I would do for some worked solutions sad

#114 Re: Help Me ! » Probability - Paper Problem » 2010-02-06 06:40:47

lindah

Thanks for your assistance!!!

I had the same line of thinking, therefore thought it couldn't be 1/4.

#115 Help Me ! » Probability - Paper Problem » 2010-02-05 09:32:16

lindah
Replies: 5

Hi all,

May I obtain any hints, tips, feedback for this question please!?!
Thank you in advance!!

There are four sheets of paper.
The first has Thai writing on one side, and an English translation on the other.
The second has Chinese writing on one side, and an English translation on the other.
The third has Korean writing on one side, and an English translation on the other.
The fourth has English writing on one side, and English on the other.
I randomly choose a sheet of paper and put in front of you, if it is English, what is the probability of the other side being English?

I am treating each side as 1 so there is a 5/8 chance of randomly picking the English side.
There is a 1/4 chance of picking the paper with English on both sides.

So using conditional probability : 1/4 divided by 5/8 = 2/5
Or is it simply 1/4? I tend to over think and confuse myself faint

#116 Re: Help Me ! » Counting unordered items » 2010-01-25 07:16:12

lindah

Hi FL,

Thanks for your prompt response
The answer is 210
Your answer made me think Hi FL, 

because I have accounted for person A & B in that position in steps 1) & 2)??

#117 Help Me ! » Counting unordered items » 2010-01-25 06:27:08

lindah
Replies: 5

Hi there,

I have an issue with the following question which I hope someone can help me with:

Nine players are to be divided into 2 teams of four and one umpire.
If two particular people cannot be on the same team together, how many different combinations are possible?

My approach was to Specify person A & B and derive the combinations

1) Person A is the umpire


2) Person B is the umpire

3) Person A is already chosen in the first team.
Person B cannot be chosen as one of the next three people, leaving

I add these combinations and get 175 - but this is incorrect.

Thanks in advance for any feedback!!

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