Answer to Question #191085 in Statistics and Probability for ANJU JAYACHANDRAN

Question #191085
If a random variable u has t -distribution with n degree of freedom, find the
distribution of u^2.
1
Expert's answer
2021-05-10T13:03:22-0400

Let s shows the sample standard deviation, μ\mu shows the population mean, xˉ\bar{x} shows the sample mean, σ\sigma is population standard deviation. and n+1 is sample size.


Chi square distribution with (n-1) degree of freedom will be


χ(n1)2s2(n1)σ2\chi^{2}_{(n-1)}\sim \frac{s^{2}(n-1)}{\sigma^{2}}


Here we will use following relationships between the variables:


Relation between standard normal distribution Z and chi-square distribution:


Z2χ12Z^{2}\sim \chi^{2}_{1}


Relation between F distribution and chi-square distribution:


Fa,bχa2/aχb2/bF_{a,b}\equiv \frac{\chi^{2}_{a}/a}{\chi^{2}_{b}/b}


The t-statistics with n degree of freedom will be


tn=xˉμs/n+1=xˉμσ/n+1σs=Zσs=Zs2σ2=Zs2(n)σ2(n)=Zχn2nt_{n}=\frac{\bar{x}-\mu}{s/\sqrt{n+1}}=\frac{\bar{x}-\mu}{\sigma/\sqrt{n+1}}\cdot \frac{\sigma}{s}=Z\cdot \frac{\sigma}{s}=\frac{Z}{\sqrt{\frac{s^{2}}{\sigma^{2}}}}=\frac{Z}{\sqrt{\frac{s^{2}(n)}{\sigma^{2}(n)}}}=\frac{Z}{\sqrt{\frac{\chi^{2}_{n}}{n}}}


 =χ121χn2n=\frac{\sqrt{\frac{\chi^{2}_{1}}{1}}}{\sqrt{\frac{\chi^{2}_{n}}{n}}}


Now squaring both sides gives:


U=tn2=χ121χn2nF1,nU=t^{2}_{n}=\frac{\frac{\chi^{2}_{1}}{1}}{\frac{\chi^{2}_{n}}{n}}\sim F_{1,n}


Hence, u2u^2 has F distribution.



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