If you seek the expectation of a function g of your random variable Z, in this case, g(Z)=exp(t(Z^2)), then compute int_{-∞}^{+∞}g(z)f(z)dz, where f(z) is the pdf of Z. The key here is that our function g is written as a function of Z, even though you see a Z^2 in there, so we use the pdf of Z to compute the expectation. Otherwise, if you wanted to let X=Z^2, then write g as a function of X, then computing the expectation would involve knowing the pdf of X, which we cannot assume when we are trying to derive the MGF of X.
Very clear demonstration. Thank you!
you are the best one on you-tube, thank you very much, sir
Why can we assume that z^2 has de same pdf as z?
If you seek the expectation of a function g of your random variable Z, in this case, g(Z)=exp(t(Z^2)), then compute int_{-∞}^{+∞}g(z)f(z)dz, where f(z) is the pdf of Z. The key here is that our function g is written as a function of Z, even though you see a Z^2 in there, so we use the pdf of Z to compute the expectation.
Otherwise, if you wanted to let X=Z^2, then write g as a function of X, then computing the expectation would involve knowing the pdf of X, which we cannot assume when we are trying to derive the MGF of X.
Thank you so much
Thank you💫
Thank you....