Proof | Property of Moment Generating Functions (5)
Proof \begin{eqnarray*}M_{aX+b}(t)&=&E[\mathrm{e}^{(aX+b)t}]\\&=&E[\mathrm{e}^{atX}\mathrm{e}^ […]
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Proof \begin{eqnarray*}M_{aX+b}(t)&=&E[\mathrm{e}^{(aX+b)t}]\\&=&E[\mathrm{e}^{atX}\mathrm{e}^ […]
もっと読む →Proof This can be proven by the property : . Actually, if we differentiate respect to , we have $$M’_{X} […]
もっと読む →Proof By the Maclaurin series expansion of , we can expand as follows: \begin{eqnarray*}M_{X}(t)&=&E[\ […]
もっと読む →Proof This is the special case of . Actually, if we differentiate respect to , we have $$M’_{X}(t)=E[X\m […]
もっと読む →Let X and Y be discrete random variables. Then the joint probability function is defined by $$f(x, y)=P(X=x, Y […]
もっと読む →Proof To proof the expression, we use the property of . If we let and , we obtain $$E[X-E[X]]=E[X]-E[X]=0$$
もっと読む →Proof Discrete Case Let f(x, y) be the joint probability function such that . \begin{eqnarray*}E[aX+bY]&=& […]
もっと読む →Proof Discrete Case Let f(x) be the probability function such that . Continuous Case Let f(x) be the probabili […]
もっと読む →Proof Discrete Case Let f(x) be the probability function such that . Continuous Case Let f(x) be the probabili […]
もっと読む →Proof : Let X be a random variable and a, b be any constants. \begin{eqnarray*}V[aX+b]&=&E[\{(aX+b)-E[ […]
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