The random variable Y = LogX has N(10,4) distribution. Find (a) The pdf of X (b) Mean and variance of X (c) P ( X ≤ 1000
for distribution N(10,4):
"\\mu=10, \\sigma^2=4"
a)
for pdf of Y:
"f_Y(y)=\\frac{1}{\\sigma\\sqrt{2\\pi}}e^{-\\frac{1}{2}(\\frac{y-\\mu}{\\sigma})^2}=\\frac{1}{2\\sqrt{2\\pi}}e^{-\\frac{1}{2}(\\frac{y-10}{2})^2}"
"\\int^{\\infin}_{-\\infin}f_Y(y)dy=1"
"dy=e^{-y}dx"
then:
"f_Y(y)=e^yf_{X>0}(x)=2e^yf_{X}(x)"
"f_{X}(x)=f_Y(y)\/(2e^y)=\\frac{1}{4x\\sqrt{2\\pi}}e^{-\\frac{1}{2}(\\frac{logx-10}{2})^2}"
b)
mean of X:
"E(X)=\\int^{\\infin}_0xf_X(x)dx=\\frac{1}{4\\sqrt{2\\pi}}\\int^{\\infin}_0e^{-\\frac{1}{2}(\\frac{logx-10}{2})^2}dx="
"=\\frac{e^{12}}{4}erf(\\frac{lnx-14}{2\\sqrt2})|^{\\infin}_0=e^{12}\/4"
where erf is the error function.
variance of X:
"Var(X)=E(X^2)-(E(X))^2"
"E(X^2)=\\int^{\\infin}_0x^2f(x)dx=\\frac{1}{4\\sqrt{2\\pi}}\\int^{\\infin}_0xe^{-\\frac{1}{2}(\\frac{logx-10}{2})^2}dx="
"=\\frac{e^{28}}{4}erf(\\frac{lnx-18}{2\\sqrt2})|^{\\infin}_0=e^{28}\/4"
"Var(X)=e^{28}\/4-e^{24}\/4"
c)
"P ( X \u2264 1000)=\\int^{1000}_0f(x)dx=\\frac{1}{4\\sqrt{2\\pi}}\\int^{1000}_0\\frac{e^{-\\frac{1}{2}(\\frac{logx-10}{2})^2}}{x}dx="
"=\\frac{1}{4}erf(\\frac{lnx-10}{2\\sqrt2})|^{1000}_0=\\frac{1}{4}(erf(\\frac{ln1000-10}{2\\sqrt2})+1)=0.0765"
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