Answer to Question #242532 in Statistics and Probability for PRS

Question #242532

Let the independent random variables X1 and X2 be N(0,1) and χ\chi 2(r) respectively. Let Y1=X1/(X2)^(1/2)/r and Y2=X2

i) Find the joint pdf of Y1 and Y2

ii) Determine the marginal pdf of Y1 and show that Y1 has a t distribution


1
Expert's answer
2022-02-01T08:42:14-0500

X1N(0,1)X_1\sim N(0,1) and its probability density function is given as,

g(x1)=12πex122, 0<x1<g(x_1)={1\over\sqrt{2\pi}}e^{-{x_1^2\over2}},\space 0\lt x_1\lt\infin

and,

X2χ2(r)X_2\sim \chi^2(r) and its probability density function is given as,

h(x2)=1Γ(r2)2r2x2r21ex22, 0<x2<h(x_2)={1\over\varGamma({r\over2})2^{r\over2}}x_2^{{r\over2}-1}e^{-{x_2\over2}},\space 0\lt x_2\lt\infin

Since the random variables X1X_1 and X2X_2 are independent, their joint probability density function is given as,

f(x1,x2)=g(x1)×h(x2)f(x_1,x_2)=g(x_1)\times h(x_2). So,

f(x1,x2)=12πex1221Γ(r2)2r2x2r21ex22, <x1<, 0<x2<0, elsewheref(x_1,x_2)={1\over\sqrt{2\pi}}e^{-{x_1^2\over2}}{1\over\varGamma({r\over2})2^{r\over2}}x_2^{{r\over2}-1}e^{-{x_2\over2}},\space -\infin\lt x_1\lt \infin,\space 0\lt x_2\lt\infin\\0,\space elsewhere

Define,

Y1=X1(X2)rY_1={X_1\over\sqrt{(X_2)\over r}} and Y2=X2Y_2=X_2. They are one to one mapping of the set, S={(x1,x2):<x1<,0<x2<}S=\{(x_1,x_2):-\infin\lt x_1\lt\infin, 0\lt x_2\lt\infin\} onto the set,

R={(y1,y2),<y1<,0<y2<}R=\{(y_1,y_2),-\infin\lt y_1\lt\infin, 0\lt y_2\lt\infin\}

So,

X2=Y2X_2=Y_2 and X1=Y1Y2r=Y1(Y2)12rX_1=Y_1\sqrt{Y_2\over r}={Y_1(Y_2)^{1\over2}\over \sqrt{r}}

The Jacobian of transformation is,

J=dx1dy1dx1dy2dx2dy1dx2dy2=y2ry12r×y201=y2rJ=\begin{vmatrix} {d_{x_1}\over d_{y_1}} & {d_{x_1}\over d_{y_2}} \\ {d_{x_2}\over d_{y_1}} & {d_{x_2}\over d_{y_2}} \end{vmatrix}=\begin{vmatrix} {\sqrt{y_2\over r}} & {y_1\over2\sqrt{r\times y_2}} \\ 0 & 1 \end{vmatrix}=\sqrt{{y_2\over r}}

Then, the joint pdf of Y1Y_1 and Y2Y_2 is given by,

g(y1,y2)=g(y1y2r,y2)J=12πΓ(r2)2r2y2r21e12y12y2rey22y2r=12πΓ(r2)2r2y2r21ey22(1+y12r)y212r, <y1<, 0<y2<g(y_1,y_2)=g(y_1\sqrt{y_2\over r},y_2)|J|\\= {1\over\sqrt{2\pi}\varGamma({r\over2})2^{r\over2}}y_2^{{r\over2}-1}e^{-{{1\over2}y_1^2}{y_2\over r}}e^{-{y_2\over2}}\sqrt{{y_2\over r}}\\= {1\over\sqrt{2\pi}\varGamma({r\over2})2^{r\over2}}y_2^{{r\over2}-1}e^{-{y_2\over2}(1+{y_1^2\over r})}{y_2^{1\over2}\over\sqrt{r}},\space -\infin\lt y_1\lt\infin,\space 0\lt y_2\lt\infin

The marginal pdf of Y1Y_1 is given by,

g(y1)=0g(y1,y2)dy2=012πΓ(r2)2r2y212(r+1)1ey22(1+y12r)y212rdy2g(y_1)=\displaystyle\int^{\infin}_0 g(y_1,y_2)dy_2=\displaystyle\int^{\infin}_0{1\over\sqrt{2\pi}\varGamma({r\over2})2^{r\over2}}y_2^{{{1\over2}(r+1)}-1}e^{-{y_2\over2}(1+{y_1^2\over r})}{y_2^{1\over2}\over\sqrt{r}}dy_2

We make the substitution,

w=y2(1+y12r)2w={y_2(1+{y_1^2\over r})\over2} and dwdy2=1+y12r2    dw=1+y12r2dy2{dw\over dy_2}={1+{y_1^2\over r}\over 2}\implies dw={1+{y_1^2\over r}\over 2}dy_2

We can now write the integral as,

012πrΓ(r2)2r2(2w1+y12r)12(r+1)1ew(21+y12r)dw\displaystyle\int_0^{\infin}{1\over\sqrt{2\pi r}\varGamma({r\over2})2^{r\over2}}({2w\over1+{y_1^2\over r}})^{{1\over2}(r+1)-1}e^{-w}({2\over1+{y_1^2\over r}})dw

Next, we introduce Γ(r+12)\varGamma({r+1\over2})

=12πrΓ(r2)2r2(21+y12r)12(r+1)Γ(r+12)0w12(r+1)1ewΓ(r+12)dw={1\over\sqrt{2\pi r}\varGamma({r\over2})2^{r\over2}}({2\over1+{y_1^2\over r}})^{{1\over2}(r+1)}\varGamma({r+1\over2})\displaystyle\int_0^{\infin}{w^{{1\over2}(r+1)-1}e^{-w}\over\varGamma({r+1\over2})}dw

The integral part is equal to 1.0 because it is a pdf of the Gamma distribution with parameters α=r+12\alpha={r+1\over2} and β=1\beta=1

So we have,

12πΓ(r2)2r2(21+y12r)12(r+1)Γ(r+12){1\over\sqrt{2\pi}\varGamma({r\over2})2^{r\over2}}({2\over1+{y_1^2\over r}})^{{1\over2}(r+1)}\varGamma({r+1\over2}) and we can write as,

g(y1)=(Γ(r+12)πrΓ(r2))(1(1+y12r)r+12), <y1<0, elsewhereg(y_1)=({\varGamma({r+1\over2})\over\sqrt{\pi r}\varGamma({r\over2})})({1\over(1+{y_1^2\over r})^{r+1\over 2}}),\space -\infin\lt y_1\lt \infin\\0,\space elsewhere

Thus, if X1N(0,1)X_1\sim N(0,1) and X2χ2(r)X_2\sim \chi^2(r) and X1X_1 and X2X_2 are independent then, the random variable Y1=X1X2rY_1={X_1\over\sqrt{X_2\over r}} has a t distribution with rr degrees of freedom.

Therefore,

Y1tdistribution(r).Y_1\sim t-distribution(r).


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