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Let \(f(t)\) denote the probability density function of \(T \sim t_{v}\). (a) Use \(f(t)\) to check that \(\mathrm{E}(T)=0, \operatorname{var}(T)=v /(v-2)\), provided \(v>1,2\) respectively. (b) By considering \(\log f(t)\), show that as \(v \rightarrow \infty, f(t) \rightarrow \phi(t)\).

Short Answer

Expert verified
(a) \(\mathrm{E}(T)=0\), \(\operatorname{var}(T)=v/(v-2)\) for \(v>2\). (b) \(f(t)\to\phi(t)\) as \(v\to\infty\).

Step by step solution

01

Define the PDF of the t-distribution

The probability density function for the t-distribution with \(v\) degrees of freedom is given by: \[ f(t) = \frac{\Gamma((v+1)/2)}{\sqrt{v\pi}\Gamma(v/2)} \left( 1 + \frac{t^2}{v} \right)^{-(v+1)/2} \] where \(\Gamma\) is the gamma function.
02

Verify the Expectation \(\mathrm{E}(T)\)

The expectation \(\mathrm{E}(T)\) of the t-distribution is zero, which is clear from the symmetry of the function \(f(t)\) about the origin. Thus, for \(v > 1\), \(\mathrm{E}(T) = 0\).
03

Verify the Variance \(\mathrm{var}(T)\)

The variance \(\mathrm{var}(T)\) of the t-distribution exists when \(v > 2\) and is given by:\[ \operatorname{var}(T) = \frac{v}{v-2} \] This is derived from the properties of the gamma function and the t-distribution.
04

Analyze \(\log f(t)\) for Large \(v\)

To show that \(\lim_{v \to \infty} f(t) = \phi(t)\), first consider the natural logarithm of \(f(t)\): \[ \log f(t) = \log \frac{\Gamma((v+1)/2)}{\sqrt{v\pi}\Gamma(v/2)} - \frac{v+1}{2} \log\left( 1 + \frac{t^2}{v} \right) \] As \(v\) approaches infinity, the terms simplify due to the properties of logarithms and \(\Gamma\) resulting in it approaching the standard normal distribution \(\phi(t)\).
05

Show Convergence to \(\phi(t)\)

The standard normal distribution \(\phi(t)\) is given by: \[ \phi(t) = \frac{1}{\sqrt{2\pi}}e^{-t^2/2} \]The t-distribution \(f(t)\) approximates \(\phi(t)\) as \(v\) increases because for large \(v\), \(\left(1 + \frac{t^2}{v}\right)^{-(v+1)/2}\) approaches \(e^{-t^2/2}\) and the gamma terms normalize similarly to \(\sqrt{2\pi}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Density Function
The probability density function (PDF) is a fundamental concept in probability theory and statistics, especially when dealing with continuous random variables. Simply put, a PDF describes the likelihood of a random variable taking on a specific value.
For the t-distribution, which is a type of probability distribution, the PDF is defined with the number of degrees of freedom, denoted by \(v\). This distribution is useful when dealing with samples that are small or have an unknown variance.

The PDF for a t-distribution is given by:
  • \( f(t) = \frac{\Gamma((v+1)/2)}{\sqrt{v\pi}\Gamma(v/2)} \left( 1 + \frac{t^2}{v} \right)^{-(v+1)/2} \)
Here, \(\Gamma\) represents the gamma function, which is a complex function that extends the factorial function to real and complex numbers. The constants and factors in this equation ensure that the area under the curve of the distribution equals 1, maintaining the property of total probability being 1.
Expectation and Variance
Expectation and variance are crucial metrics in statistics to understand the distribution of data. Expectation, \(\mathrm{E}(T)\), is the average or mean value expected from repeated trials of an experiment. For a t-distribution, this mean (or expectation) is zero, represented by \(\mathrm{E}(T) = 0\). This zero expectation is due to the symmetry of the t-distribution about the y-axis, provided that the degrees of freedom \(v > 1\).

Variance, represented as \(\operatorname{var}(T)\), measures how spread out the values of the random variable are around the mean. For the t-distribution:
  • The variance is computed as \(\operatorname{var}(T) = \frac{v}{v-2}\) for \(v > 2\).
This variance is greater than 1, illustrating that t-distributions have heavier tails than the normal distribution, meaning they allow for a higher likelihood of values further from the mean.
Gamma Function
The gamma function (\(\Gamma(x)\)) plays a vital role in mathematics and statistics, particularly in defining distributions like the t-distribution. It generalizes the factorial function to non-integer values. While \(n!\) is defined as the product of all positive integers up to \(n\), \(\Gamma(x)\) is defined for real and complex numbers and extends this factorial concept.

The gamma function appears in the t-distribution formula, helping to define the shape and probability of the distribution. The function can be written as:
  • \(\Gamma(n) = (n-1)!\) if \(n\) is a positive integer.
  • For a non-integer, the gamma function is defined by an integral: \(\Gamma(x) = \int_0^\infty t^{x-1}e^{-t} \, dt\).
The gamma function ensures that the PDF is correctly normalized, so all probability sums up to 1.
Convergence to Normal Distribution
The concept of convergence to a normal distribution refers to how a series of distributions stabilizes to resemble the normal distribution as certain parameters approach infinity.
For the t-distribution, this convergence becomes evident as the degrees of freedom \(v\) increase. As \(v\) grows larger, the shape of the t-distribution becomes more like a standard normal distribution \(\phi(t)\), defined as:
  • \(\phi(t) = \frac{1}{\sqrt{2\pi}}e^{-t^2/2}\)
This transformation is crucial as it allows for the use of normal distribution approximations in statistical analysis when sample sizes are large. As the degrees of freedom increase, the t-distribution's heavy tails (which indicate higher probabilities of extreme values) diminish, and it starts matching the bell curve characteristic of \(\phi(t)\), simplifying many practical statistical calculations.

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Most popular questions from this chapter

I am uncertain about what will happen when I next roll a die, about the exact amount of money at present in my bank account, about the weather tomorrow, and about what will happen when I die. Does uncertainty mean the same thing in all these contexts? For which is variation due to repeated sampling meaningful, do you think?

Independent pairs \(\left(X_{j}, Y_{j}\right), j=1, \ldots, m\) arise in such a way that \(X_{j}\) is normal with mean \(\lambda_{j}\) and \(Y_{j}\) is normal with mean \(\lambda_{j}+\psi, X_{j}\) and \(Y_{j}\) are independent, and each has variance \(\sigma^{2} .\) Find the joint distribution of \(Z_{1}, \ldots, Z_{m}\), where \(Z_{j}=Y_{j}-X_{j}\), and hence show that there is a \((1-2 \alpha)\) confidence interval for \(\psi\) of form \(A \pm m^{-1 / 2} B c\), where \(A\) and \(B\) are random variables and \(c\) is a constant. Obtain a \(0.95\) confidence interval for the mean difference \(\Psi\) given \((x, y)\) pairs \((27,26)\), \((34,30),(31,31),(30,32),(29,25),(38,35),(39,33),(42,32) .\) Is it plausible that \(\psi \neq 0 ?\)

Let \(R_{1}, R_{2}\) be independent binomial random variables with probabilities \(\pi_{1}, \pi_{2}\) and denominators \(m_{1}, m_{2}\), and let \(P_{i}=R_{i} / m_{i} .\) It is desired to test if \(\pi_{1}=\pi_{2}\). Let \(\widehat{\pi}=\left(m_{1} P_{1}+m_{2} P_{2}\right) /\left(m_{1}+m_{2}\right) .\) Show that when \(\pi_{1}=\pi_{2}\), the statistic $$ Z=\frac{P_{1}-P_{2}}{\sqrt{\widehat{\pi}(1-\hat{\pi})\left(1 / m_{1}+1 / m_{2}\right)}} \stackrel{D}{\longrightarrow} N(0,1) $$ when \(m_{1}, m_{2} \rightarrow \infty\) in such a way that \(m_{1} / m_{2} \rightarrow \xi\) for \(0<\xi<1\). Now consider a \(2 \times 2\) table formed using two independent binomial variables and having entries \(R_{i}, S_{i}\) where \(R_{i}+S_{i}=m_{i}, R_{i} / m_{i}=P_{i}\), for \(i=1,2\). Show that if \(\pi_{1}=\pi_{2}\) and \(m_{1}, m_{2} \rightarrow \infty\), then $$ X^{2}=\left(n_{1}+n_{2}\right)\left(R_{1} S_{2}-R_{2} S_{1}\right)^{2} /\left\\{n_{1} n_{2}\left(R_{1}+R_{2}\right)\left(S_{1}+S_{2}\right)\right\\} \stackrel{D}{\longrightarrow} \chi_{1}^{2} $$ Two batches of trees were planted in a park: 250 were obtained from nursery \(A\) and 250 from nursery \(B\). Subsequently 41 and 64 trees from the two groups die. Do trees from the two nurseries have the same survival probabilities? Are the assumptions you make reasonable?

If \(Z\) is standard normal, then \(Y=\exp (\mu+\sigma Z)\) is said to have the log-normal distribution. Show that \(\mathrm{E}\left(Y^{r}\right)=\exp (r \mu) M_{Z}(r \sigma)\) and hence give expressions for the mean and variance of \(Y\). Show that although all its moments are finite, \(Y\) does not have a moment- generating function.

\(W_{i}, X_{i}, Y_{i}\), and \(Z_{i}, i=1,2\), are eight independent, normal random variables with common variance \(\sigma^{2}\) and expectations \(\mu_{W}, \mu_{X}, \mu_{Y}\) and \(\mu_{Z} .\) Find the joint distribution of the random variables $$ \begin{aligned} T_{1} &=\frac{1}{2}\left(W_{1}+W_{2}\right)-\mu_{W}, T_{2}=\frac{1}{2}\left(X_{1}+X_{2}\right)-\mu_{X} \\ T_{3} &=\frac{1}{2}\left(Y_{1}+Y_{2}\right)-\mu_{Y}, T_{4}=\frac{1}{2}\left(Z_{1}+Z_{2}\right)-\mu_{Z} \\ T_{5} &=W_{1}-W_{2}, T_{6}=X_{1}-X_{2}, T_{7}=Y_{1}-Y_{2}, T_{8}=Z_{1}-Z_{2} \end{aligned} $$ Hence obtain the distribution of $$ U=4 \frac{T_{1}^{2}+T_{2}^{2}+T_{3}^{2}+T_{4}^{2}}{T_{5}^{2}+T_{6}^{2}+T_{7}^{2}+T_{8}^{2}} $$ Show that the random variables \(U /(1+U)\) and \(1 /(1+U)\) are identically distributed, without finding their probability density functions. Find their common density function and hence determine \(\operatorname{Pr}(U \leq 2)\).

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