Suppose X1 and X2 are a random sample from the following...

Suppose $X_1$ and $X_2$ are a random sample from the following distribution: \begin{equation} f(x) = \frac{1}{\Gamma(\alpha)\beta^{\alpha}}x^{\alpha-1}e^{-x/\beta}, \quad 0 < x < \infty, \quad 0 < \alpha, \beta < \infty. \end{equation} (a) Find the joint pdf of $Y_1 = X_1/(X_1 + X_2)$ and $Y_2 = X_1 + X_2$. (b) Find $E(Y_1^2)$.

Transcript

00:01 So we will find the joint probability of density functions of y1 and y2, so we will use a method of transformation of random variable.

00:12 So transformation are given, so y1 is equal to x1 square and y2 is equal to x1 multiplied by x2.

00:41 So we need find the joint pdf y1 is equal to small y1, y2 is equal to small y2.

00:56 So what we express, so find to need to determine the relationship of x1 and x2 and y1 and y2, there is a jhokha -dinh determine.

01:14 So what he determine, he has y1 is equal to x1 square and y2 is equal to x1 multiplied by x2.

01:34 So these are 1 value and 2 value.

01:44 So the inverse transformation, so inverse transformation is x1 and x2 is equal to square root of y1 and y square divided by 1.

02:17 So we find the jhokha -dinh determinant, j.

02:31 So we find the j, so how we find the j? so we see j is equal to inverse x1 divided by x, so divided by y1, so multiplied by x2, again y1, so write j.

03:25 Then again multiplied by x2, no multiplied by these 2 value and then minus x1 y2, x2 y2.

03:54 So only write, so i think i write again, so i write again j is equal to x1 divided by y1, x2 y1, x1, so y2 y2 x2.

04:51 So this one is multiplied by minus multiplied by, so bracket, put bracket here.

05:13 So this is a j, jhokha -dinh determinant.

05:19 So the first one value x1 divided by y1 is equal to 1.

05:30 So 2 multiplied by square root of y1, so the value is equal to 1.

05:38 So x2 y1 is equal to 0.

05:45 So x1 y2, the value is equal to minus y2.

05:56 So 2 multiplied by y1, so 3 divided by 2, the answer is minus y2.

06:09 So the last one is x1, y, x2 and y2.

06:16 So the value is 1 divided by square root of y1.

06:28 So putting all the values in j, so 1 divided by 2 multiplied by square root of y1.

06:45 So multiplied by 1 divided by square root of y1.

07:20 So minus 0 multiplied by minus y square divided by 2 y3 divided by 2.

07:38 So j value is equal to 1 divided by 2 multiplied by y1.

07:50 So then we express to giant a pdf y1 and y2.

08:03 So giant pdf of x1 and x2.

08:06 So f is equal to y1, y2 is equal to f x1 x2 multiplied by j...

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