Show directly from the pdf that the mean of a t_1 (Cauchy) random variable does

Question

Show directly from the pdf that the mean of a t_1 (Cauchy) random variable does not exist. Show that the ratio of two independent standard normal random variables has the t_1 distribution. Apply the method used to derive the t pdf in this section. Let X have an F distribution with v_1 numerator df and v_2 denominator df. Determine the mean value of X. Determine the variance of X. Is it true that D(F_v_1, v_2) = E(chi_v_1^2/v_1)/E(chi_v_2^2/v_2) Explain. Show that F_p, v_1, v_2 = 1/F_1-p, v_2, v_1. Derive the F pdf by applying the method used to derive the t pdf.

Explanation / Answer

True,

Since chi square test statistic is used to check whether the single variance is significant or not.

2 = (v-1) S2 / 2

If we take two chi square test statistics for each variance tests individually

then its ratio of two 2 variates with v1 and v2 df follows F distribution with (v1, v2) df

Here F statistic = S12 / S22

  Now its means are equal

Therefore E(F) = E(21) / E(22)

Given statement is true

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