Using the notation in this chapter, prove that a 95 % confidence interval for ST is between S0 e

Using the notation in this chapter, prove that a 95 %...

Key Concepts

Confidence Interval

A confidence interval is a range of values derived from sample data that is believed, with a certain probability (in this case, 95%), to contain the true parameter of interest. It reflects the uncertainty inherent in any estimate obtained from random data. In statistical inference, the interval is constructed such that if the experiment were repeated many times, the observed interval would capture the true parameter value a specified percentage of the time.

Normal Distribution

The normal distribution is a foundational probability distribution in statistics characterized by its bell-shaped curve. It is defined by its mean and standard deviation and has well-known properties, such as symmetry around the mean and the empirical rule for standard deviations. When dealing with sums or averages of random variables—as in the context of log returns—the normal distribution is often applicable due to the central limit theorem.

Lognormal Distribution

A lognormal distribution arises when the logarithm of a random variable is normally distributed. This is particularly important in financial modeling and other areas where a variable, such as stock prices, is always positive and evolves multiplicatively. In the given context, because the logarithm of the asset price follows a normal distribution, the asset price itself is lognormally distributed, which influences how confidence intervals are constructed for it.

Transformation of Variables

The transformation of variables involves converting a random variable under one distribution into another form, often through functions such as the logarithm or the exponential. In this case, by taking the logarithm of the asset price, which follows a lognormal distribution, one obtains a normally distributed variable. The confidence interval for the original variable is then derived by first obtaining the interval on the log scale and subsequently applying the exponential function to restore the original scale.

Transcript

00:02 Hi, here it is given that x1, x2, up to xn, these are n random samples which are iid follows normal distribution with mean mu 1 and variance sigma 1 square.

00:19 Similarly y1, y2, another set of random variables.

00:25 Better x1 x2 we can take n1 and this y n2.

00:29 Okay our i iid this follows normal distribution with mean mu two and variance sigma 2 okay now we s 1 square we denote is the sample variance of this x1 x2 xx this is equals to 1 by n1 minus 1 summation irons from 1 to n1 x i minus xx 4 whole square x bar is the sample mean 1 by n1 summation x i runs from 1 to n1 okay and h2 square is the sample variance of this y 1 y 2 y n 2 this is 1 by n 2 summation j runs from 1 to n2 n2 n2 into y j minus y bar square where y j is the sample mean of y 1 y 2 y n 2 1 1 2 1 1 1 2 1 1 1 and to summation j runs from 1 to n to y j now what we have to show that root over s1 square whole divided by n1 plus s2 square whole divided by n2 whole divided by n2 whole divided by root over sigma 1 square whole divided by n1 plus sigma 2 whole divided by n2, this converges in probability to 1 as n1 n2 tends to infinity.

02:15 Okay, let's start with the problem.

02:17 See as here xi follows normal mu 1 sigma 1 square distribution so we can write that s1 square equals to 1 by n1 minus 1 summation irons from 1 to n1 x i .m minus x bar square so from here it can be written that summation irons from 1 to n1 x i .i minus x bar world developed by sigma 1 square this follows kai square distribution with parameter n1 minus 1 right so this means n1 minus 1 into s 1 square whole divided by sigma 1 square follows k -square distribution with parameter n -1 minus 1 so from here we can write expectation of n -1 minus 1 n 1 into s 1 square whole divided by sigma 1 square this is equals to n -1 minus 1 because mean of kai -scar distribution is n -1 minus 1 and variance is to 2 into n1 minus 1 okay so we get expectation of s 1 square equals to sigma 1 square now variance of variance of n1 minus 1 s 1 square hold by sigma 1 square this is equals to 2 into n1 minus 1 right from here we get variance of s1 square equals to 2 into sigma 1 to 4 whole divided by n1 minus 1...

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