Given information
In the question we are given with the following two statements:
Statement 1: "The means of the Student’s t and standard normal distributions are equal."
Statement 2: "The standard normal distribution approaches the Student’s t distribution as the degrees of freedom becomes large."
and we have to check whether they are true or not.
The probability density function (pdf) of standard normal distribution is:
\(\displaystyle{{f}_{{X}}{\left({x}\right)}}={\int_{{-\infty}}^{{\infty}}}\frac{{1}}{{\sqrt{{{2}\pi}}}}{\exp{{\left\lbrace\frac{{-{x}^{{2}}}}{{{2}}}\right\rbrace}}}{\left.{d}{x}\right.}\)
where X is a random variable following Standard Normal distribution.And, standard normal distribution have mean as 0 and variance as 1.
\(\displaystyle{X}\sim{N}{\left({0},{1}\right)}\)
The probability density function of student's t distribution is:
\(\displaystyle{{f}_{{X}}{\left({x}\right)}}={\int_{{-\infty}}^{{\infty}}}\frac{{\Gamma{\left({v}+{1}\right)}}}{{\sqrt{{{v}\pi}}\Gamma{\left(\frac{{v}}{{2}}\right)}}}{\left({1}+\frac{{{x}^{{2}}}}{{v}}\right)}^{{{v}+{1}}}\)
where X is a random variable following student's t distribution with mean 0 and variance \(\displaystyle\frac{{v}}{{{v}-{2}}}\).
\(\displaystyle{X}\sim{t}_{{v}}\)
Explanation
The statement 1 is true as both student's t, and standard normal distribution have mean equal to zero.
The statement 2 is false as in case of large sample size, the student's t distribution approaches to the standard normal distribution, or it can be said that when the degree of freedom is large, the student's t distribution approaches to standard normal distribution instead of standard normal distribution approaches to student's t with a large degree of freedom.
Hence, the answer is (A) I only.
In the question we are given with the following two statements:
Statement 1: "The means of the Student’s t and standard normal distributions are equal."
Statement 2: "The standard normal distribution approaches the Student’s t distribution as the degrees of freedom becomes large."
and we have to check whether they are true or not.
The probability density function (pdf) of standard normal distribution is:
\(\displaystyle{{f}_{{X}}{\left({x}\right)}}={\int_{{-\infty}}^{{\infty}}}\frac{{1}}{{\sqrt{{{2}\pi}}}}{\exp{{\left\lbrace\frac{{-{x}^{{2}}}}{{{2}}}\right\rbrace}}}{\left.{d}{x}\right.}\)
where X is a random variable following Standard Normal distribution.And, standard normal distribution have mean as 0 and variance as 1.
\(\displaystyle{X}\sim{N}{\left({0},{1}\right)}\)
The probability density function of student's t distribution is:
\(\displaystyle{{f}_{{X}}{\left({x}\right)}}={\int_{{-\infty}}^{{\infty}}}\frac{{\Gamma{\left({v}+{1}\right)}}}{{\sqrt{{{v}\pi}}\Gamma{\left(\frac{{v}}{{2}}\right)}}}{\left({1}+\frac{{{x}^{{2}}}}{{v}}\right)}^{{{v}+{1}}}\)
where X is a random variable following student's t distribution with mean 0 and variance \(\displaystyle\frac{{v}}{{{v}-{2}}}\).
\(\displaystyle{X}\sim{t}_{{v}}\)
Explanation
The statement 1 is true as both student's t, and standard normal distribution have mean equal to zero.
The statement 2 is false as in case of large sample size, the student's t distribution approaches to the standard normal distribution, or it can be said that when the degree of freedom is large, the student's t distribution approaches to standard normal distribution instead of standard normal distribution approaches to student's t with a large degree of freedom.
Hence, the answer is (A) I only.
