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...e following system: Y=a+bX+cX^2+dX^3, where X~N(0,1). (Y is expressed as a linear combination of the first three powers of a standard normal variable.) Assuming that E(Y)=0 and Var(Y)=1, one can obtain the following equations after tedious algebraic calculations: 1) b^2+6bd+2c^2+15d^2=1 2) 2c(b^2+24bd+105d^2+2)=E(Y^3) 3) 24[bd+c^2(1+b^2+28bd)+d^2(12+48bd+141c^2+225d^2)]=E(Y^4)-3 Obviously, a=-c. Suppose that distributional form of Y is given so we know E(Y^3) and E(Y^4). In other words, we have access to the third and fourth raw moments. How do we solve for these four coefficients? I reduced th...