Good morning sir, just a question. In the example for the transformation from a normal distribution to a standard normal distribution, why would x be 11 and not 10*(1.11) 0 11.1? The question says what is the probability that the return will be less than 11%. So, if we do Z = [(10*1.11) - 10] / 2 0 0.55, and the z value is obviously different. Can you please explain? Best, Marco
for Portfolio mean = (wA*µA) + (wB*µB) + (wC*µC) σP = (wA^2σA^2 + wB^2 σB^2 + wC^2σC^2 + 2wAwBσAσBρAB + 2wBwCσBσCρBC + 2wAwCσAσCρAC)^1/2 use the above formulas. Since the complete information is not provided (i.e. returns of stock A, B, C) in the example. The example cannot be solved. IFT support team
what is the difference between 95% data being in 2 standard deviations from the mean vs, 95% confidence interval that 95% of data being 1.96 standard deviations from the mean? or 99% data being in 3 standard deviation from the mean vs, 99% confidence interval that 99% of data being 2.58 standard deviation from the mean? Thank you.
Approximately 95% of all observations fall in the interval m ± 2s. whereas, 95% of all observations are in the interval m ± 1.96s. The key difference is “approximation”. IFT support team
Chebyshev’s inequality says that at least 1-1/K2 of data from a sample must fall within K standard deviations from the mean (here K is any positive real number greater than one). Any data set that is normally distributed, or in the shape of a bell curve, has several features. One of them deals with the spread of the data relative to the number of standard deviations from the mean. In a normal distribution, we know that 68% of the data is one standard deviation from the mean, 95% is two standard deviations from the mean, and approximately 99% is within three standard deviations from the mean. But if the data set is not distributed in the shape of a bell curve, then a different amount could be within one standard deviation. Chebyshev’s inequality provides a way to know what fraction of data falls within K standard deviations from the mean for any data set. IFT Support Team
which example are your referring to in this lecture? Generally, the most frequently cited facts that result from Chebyshev’s inequality are that a two- standard- deviation interval around the mean must contain at least 75 percent of the observations, and a three- standard- deviation interval around the mean must contain at least 89 percent of the observations, no matter how the data are distributed. IFT Support Team
Dear Sirs, could you please explain how did you get to 0.5-0.31 in the example of using z-table? I get to 0,19 by the following: (1-(0,31*2))/2. But as I can see your calculations is more simple.
Dear Анатолий Идзиковский, Firstly we calculate the z-value= (11-10)/2 = 0.5. Then we look up for 0.5 on z-table which gives us the probability of 0.6915, this is the probability to the left of 0.5(or probability of less than 11%). Then to calculate the probability of being on the right of 0.5(probability of greater than 11%) we do 1-0.69=0.31. IFT Support Team
A normally distributed random variable has a mean of 100 and a standard deviation of 12. The probability of observing a value grater than 82 is the cumulative distribution (cdf) of the standard normal variable: N(1.5). Why is this the correct answer? Should it be "1- N(1.5)" since "N(1.5)" provides the probability of less than or equal to 85? Thank you for helping me with that sir, I really appreciate your video.
The standardized value of this normal distribution can be obtained using the formula =(X-μ)/σ =(82-100)/12=- 1.5. The cdf of N(-1.5) provides the probability of a value less than or equal to 82. Probability of a value smaller than 82= N(-1.5) b/c if symmetry this is identical to 1-N(1.5). Now we want to know the probability of observing a value greater than 82, which would then be: Probability of a value bigger than 82= 1-N(-1.5) b/c if symmetry this is identical to 1-(1-N(1.5))=N(1.5) IFT support team
Dear Jean, No. However, you only need to know a few important value, which are mention in our videos (Like z value for 95% confidence interval). IFT Support Team
Whats difference within confidence interval and Chebyshev's inequality other than Chebyshev's inequality applies to all kind of distributions and confidence interval applies only for normal distribution?
A confidence interval is a range for which one can assert with a given probability 1 − α, called the degree of confidence, that it will contain the parameter it is intended to estimate.According to Chebyshev’s inequality, for any distribution with finite variance, the proportion of the observations within k standard deviations of the arithmetic mean is at least 1 − 1/k2 for all k > 1. Source: CFAI Curriculum.When k = 1.25, for example, the inequality states that the minimum proportion of the observations that lie within ± 1.25s is 1 − 1/(1.25)2 = 1 − 0.64 = 0.36 or 36 percent.IFT Support Team
hello Sir, i would like to ask again the same question , during exam how will we find the values quickly of the table, because in the video you quickly referred to the table , if cfa doesn't provide us the same , please let us know because i cant figure out the calculation Thanks
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This can’t be more simple 😊
Sir, For problem sum to find eps between 4 and 5... Shouldn't we subtract 0.69-0.31 why have we done 0.50-0.31. Thanks
No. The area till 0 is 0.5. Hence, we will do 0.5 - 0.31.
IFT Support Team
Good morning sir, just a question. In the example for the transformation from a normal distribution to a standard normal distribution, why would x be 11 and not 10*(1.11) 0 11.1? The question says what is the probability that the return will be less than 11%. So, if we do Z = [(10*1.11) - 10] / 2 0 0.55, and the z value is obviously different.
Can you please explain?
Best,
Marco
We take 11 because X is 11 here. 10 is mean return.
IFT Support Team
Dear Sir
Do we need to memorize the probablity (x or z) table for the examination?
Thank you!
the probability table will be provided in the exam. However, you may memorize the important ones, i.e. 90%, 95%, 99%.
IFT Support Team
Hey!
Will we get the x and z table for probability during the exam?
Can i know what will be the answer for the final example discussed or the method to solve it, I am referring to the Multivariate Distribution example
for Portfolio mean = (wA*µA) + (wB*µB) + (wC*µC)
σP = (wA^2σA^2 + wB^2 σB^2 + wC^2σC^2 + 2wAwBσAσBρAB + 2wBwCσBσCρBC + 2wAwCσAσCρAC)^1/2
use the above formulas. Since the complete information is not provided (i.e. returns of stock A, B, C) in the example. The example cannot be solved.
IFT support team
what is the difference between 95% data being in 2 standard deviations from the mean vs, 95% confidence interval that 95% of data being 1.96 standard deviations from the mean?
or 99% data being in 3 standard deviation from the mean vs, 99% confidence interval that 99% of data being 2.58 standard deviation from the mean? Thank you.
Approximately 95% of all observations fall in the interval m ± 2s. whereas, 95% of all observations are in the interval m ± 1.96s. The key difference is “approximation”.
IFT support team
Is there any connection between Chebeshev's equality & confidence intervals for a normal distribution?
Chebyshev’s inequality says that at least 1-1/K2 of data from a sample must fall within K standard deviations from the mean (here K is any positive real number greater than one). Any data set that is normally distributed, or in the shape of a bell curve, has several features. One of them deals with the spread of the data relative to the number of standard deviations from the mean. In a normal distribution, we know that 68% of the data is one standard deviation from the mean, 95% is two standard deviations from the mean, and approximately 99% is within three standard deviations from the mean. But if the data set is not distributed in the shape of a bell curve, then a different amount could be within one standard deviation. Chebyshev’s inequality provides a way to know what fraction of data falls within K standard deviations from the mean for any data set.
IFT Support Team
I find 68% figure is violating the chebyshev's formula which suggest minimum 75% should lie for +-2 deviation. Where am I wrong?
which example are your referring to in this lecture? Generally, the most frequently cited facts that result from Chebyshev’s inequality are that a two- standard- deviation interval around the mean must contain at least 75 percent of the observations, and a three- standard- deviation interval around the mean must contain at least 89 percent of the observations, no matter how the data are distributed.
IFT Support Team
Dear Sirs, could you please explain how did you get to 0.5-0.31 in the example of using z-table? I get to 0,19 by the following: (1-(0,31*2))/2. But as I can see your calculations is more simple.
Dear Анатолий Идзиковский,
Firstly we calculate the z-value= (11-10)/2 = 0.5. Then we look up for 0.5 on z-table which gives us the probability of 0.6915, this is the probability to the left of 0.5(or probability of less than 11%). Then to calculate the probability of being on the right of 0.5(probability of greater than 11%) we do 1-0.69=0.31.
IFT Support Team
22:15 and 15:30
is it possible to get a hand on all of the slides that you have used in all of your quants videos together?
Sir, is mean for a normal distribution always 10 and standard deviation 2?
it is just an example
IFT support team
A normally distributed random variable has a mean of 100 and a standard deviation of 12. The probability of observing a value grater than 82 is the cumulative distribution (cdf) of the standard normal variable: N(1.5).
Why is this the correct answer? Should it be "1- N(1.5)" since "N(1.5)" provides the probability of less than or equal to 85?
Thank you for helping me with that sir, I really appreciate your video.
The standardized value of this normal distribution can be obtained using the formula =(X-μ)/σ =(82-100)/12=- 1.5. The cdf of N(-1.5) provides the probability of a value less than or equal to 82. Probability of a value smaller than 82= N(-1.5) b/c if symmetry this is identical to 1-N(1.5). Now we want to know the probability of observing a value greater than 82, which would then be: Probability of a value bigger than 82= 1-N(-1.5) b/c if symmetry this is identical to 1-(1-N(1.5))=N(1.5)
IFT support team
Hey, would the standard normal distribution table be given on the day of the exam? Please! I would like to know.
Dear Jean,
No. However, you only need to know a few important value, which are mention in our videos (Like z value for 95% confidence interval).
IFT Support Team
Thank you so much.. This is very helpful.
Glad it was helpful!
IFT support team
Whats difference within confidence interval and Chebyshev's inequality other than Chebyshev's inequality applies to all kind of distributions and confidence interval applies only for normal distribution?
A confidence interval is a range for which one can assert with a given probability 1 − α, called the degree of confidence, that it will contain the parameter it is intended to estimate.According to Chebyshev’s inequality, for any distribution with finite variance, the proportion of the observations within k standard deviations of the arithmetic mean is at least 1 − 1/k2 for all k > 1.
Source: CFAI Curriculum.When k = 1.25, for example, the inequality states that the minimum proportion of the observations that lie within ± 1.25s is 1 − 1/(1.25)2 = 1 − 0.64 = 0.36 or 36 percent.IFT Support Team
chebeshev's statement is atleast and confidence interval is % surety
hello Sir, i would like to ask again the same question , during exam how will we find the values quickly of the table, because in the video you quickly referred to the table , if cfa doesn't provide us the same , please let us know because i cant figure out the calculation
Thanks
You only need to know a few important value, which are mention in our videos (Like z value for 95% confidence interval).
IFT Support Team
IFT 0