A continuous random variable X is said to have a Normal Distribution if its PDF is:
"f_X(x)=\\frac{1}{\u03c3 \\sqrt{2\u03c0}}e^{\u2212\\frac{1}{2}(\\frac{x\u2212\u03bc}{\u03c3})^2},x\u2208R\\\\"
So if X∼N(μ,σ2) then it’s MGF is:
"M_x(t)\\\\ \n\n=E(e^{tx})\\\\\n\n=\u222b^\u221e_{\u2212\u221e}e^{tx}\\frac{1}{\u03c3 \\sqrt{2\u03c0}}e^{\u2212\\frac{1}{2}}(\\frac{x\u2212\u03bc}{\u03c3})^2dx\\\\\n\n=\u222b^\u221e_{\u2212\u221e}e^t(\u03bc+\u03c3z)\\frac{1}{\\sqrt{2\u03c0}}e\u2212\\frac{1}{2}z^2dz, where \\frac{x\u2212\u03bc}{\u03c3}=z\\\\\n\n=(e^{t\u03bc}+\\frac{1}{2}t^2\u03c3^2)\u222b^\u221e_{\u2212\u221e}\\frac{1}{\\sqrt{2\u03c0}}e^\u2212\\frac{1}{2}(z\u2212t\u03c3)^2dz,"
Note that the portion under integration sign represents PDF of N(tσ,1) distribution; so eventually the value of the integration will be 1
"e^{t\u03bc}+\\frac{1}{2}t^2\u03c3^2, t \u2208R"
So, "M_x(t)= e^{t\u03bc}+\\frac{1}{2}t^2\u03c3^2, t \u2208R"
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