Consider random variables S: Ohm rightarrow R and T: Ohm rightarrow R, where Ohm

Question

Consider random variables S: Ohm rightarrow R and T: Ohm rightarrow R, where Ohm Subsetequalto R^n. Let P be a probability measure on Ohm with density g(x). One way to compute a conditional expectation E[S|T = t] is to first perform a change of variables (y_1, ..., y_n) = phi (x_1, ..., x_n) where y_n = T (x_1, ..., x_n) and phi: Ohm rightarrow Ohm' is differentiable and invertible, with Ohm' Subsetequalto R^n. Then E[S|T = t] = integral_Ohm" S(x)|J(x)|^-1 g(x) dy_1 ... dy_n - 1/integral_Ohm" |J(x)|^-1 g(x) dy_1 ... dy_n - 1 where x = phi^-1 (y), |J(x)| is the Jacobian determinant for phi and Ohm" = {y elementof R^n - 1: (y_1, ..., y_n - 1, t) elementof Ohm'}. Let X_1, ..., X_n be independent and identically distributed exponential random variables with common density f(x_i|theta) = theta e^-theta x_i with x_i > 0 for i = 1, ..., n. Use the change of variables technique to evaluate E[X^2_1|T = t], where T = sigma^n_i = 1 X_i.

Explanation / Answer

we can obtain the pdf by applying the change of variable technique to the following pdf of the random vector WW

fW(w)=i=1kewiwi1i/(i),fW(w)=i=1kewiwii1/(i),

and the invertible transformation W(X1,…,Xk1,Z)W(X1,…,Xk1,Z) where Xi=Wiki=1WiXi=Wii=1kWi and Z=ki=1WiZ=i=1kWi. We can express the transformation as Wi=XiZWi=XiZ for i=1,…,k1i=1,…,k1 and Wk=ZXk=Z(1X1Xk1)Wk=ZXk=Z(1X1Xk1). It follows that the Jacobian of the transformation is the following matrix

J=Z0Z0ZZX1X2Xk.J=(Z0X10ZX2ZZXk).

To use the change of variables technique, we need to compute the determinant of the matrix above. Recalling that adding one of the rows of the matrix to another row does not change the determinant, we can add the first k1k1 rows to the last row, and then compute the determinant of the resulting matrix. That matrix is an upper diagonal matrix (all elements below the diagonal are zero) with diagonal elements Z,Z,,Z,1Z,Z,,Z,1. Since the determinant of an upper diagonal matrix is the product of the diagonal terms (all other terms in the product defining the determinant expression zero out), the determinant is Zk1Zk1. It follows that the pdf of the transformed variables (X1,…,Xk1,Z)(X1,…,Xk1,Z) is

fX1,…,Xk1,Z(x1,…,xk1,z)=zk1i=1kexiz(xiz)i1/(i)=zi1e1zi=1kxi1i/(i).fX1,…,Xk1,Z(x1,…,xk1,z)=zk1i=1kexiz(xiz)i1/(i)=zi1e1zi=1kxii1/(i).

We integrate the above expression with respect to zz in order to obtain the pdf of X1,…,Xk1X1,…,Xk1.

fX1,…,Xk1(x1,…,xk1)=i=1kxi1i/(i)0zi1e1zdz=i=1kxi1i/(i)(i=1ki)fX1,…,Xk1(x1,…,xk1)=i=1kxii1/(i)0zi1e1zdz=i=1kxii1/(i)(i=1ki)

where the last equation follows from the fact that Gamma pdf integrates to one (see Section 3.10). Since ki=1Xi=1i=1kXi=1, the RV XkXk must equal 1k1i=1Xi1i=1k1Xi. The pdf above therefore remains the same if we add the RV XkXk taking value 1k1i=1Xi1i=1k1Xi (the pdf becomes zero for all other values of XkXk).

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