Let the independent random variables X1 and X2 be N(0,1) and "\\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
"X_1\\sim N(0,1)" and its probability density function is given as,
"g(x_1)={1\\over\\sqrt{2\\pi}}e^{-{x_1^2\\over2}},\\space 0\\lt x_1\\lt\\infin"
and,
"X_2\\sim \\chi^2(r)" and its probability density function is given as,
"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 "X_1" and "X_2" are independent, their joint probability density function is given as,
"f(x_1,x_2)=g(x_1)\\times h(x_2)". So,
"f(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,
"Y_1={X_1\\over\\sqrt{(X_2)\\over r}}" and "Y_2=X_2". They are one to one mapping of the set, "S=\\{(x_1,x_2):-\\infin\\lt x_1\\lt\\infin, 0\\lt x_2\\lt\\infin\\}" onto the set,
"R=\\{(y_1,y_2),-\\infin\\lt y_1\\lt\\infin, 0\\lt y_2\\lt\\infin\\}"
So,
"X_2=Y_2" and "X_1=Y_1\\sqrt{Y_2\\over r}={Y_1(Y_2)^{1\\over2}\\over \\sqrt{r}}"
The Jacobian of transformation is,
"J=\\begin{vmatrix}\n {d_{x_1}\\over d_{y_1}} & {d_{x_1}\\over d_{y_2}} \\\\\n {d_{x_2}\\over d_{y_1}} & {d_{x_2}\\over d_{y_2}}\n\\end{vmatrix}=\\begin{vmatrix}\n {\\sqrt{y_2\\over r}} & {y_1\\over2\\sqrt{r\\times y_2}} \\\\\n 0 & 1\n\\end{vmatrix}=\\sqrt{{y_2\\over r}}"
Then, the joint pdf of "Y_1" and "Y_2" is given by,
"g(y_1,y_2)=g(y_1\\sqrt{y_2\\over r},y_2)|J|\\\\=\n{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}}\\\\=\n{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 "Y_1" is given by,
"g(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={y_2(1+{y_1^2\\over r})\\over2}" and "{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,
"\\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 "\\varGamma({r+1\\over2})"
"={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 "\\alpha={r+1\\over2}" and "\\beta=1"
So we have,
"{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(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 "X_1\\sim N(0,1)" and "X_2\\sim \\chi^2(r)" and "X_1" and "X_2" are independent then, the random variable "Y_1={X_1\\over\\sqrt{X_2\\over r}}" has a t distribution with "r" degrees of freedom.
Therefore,
"Y_1\\sim t-distribution(r)."
Comments
Leave a comment