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Question Number 6604 by WAI LIN last updated on 05/Jul/16
If the density function of X equals  f(x) =  { ((c e^(−2x)  , 0<x<∞)),((0 ,          x<0)) :}  find c. What is P{X>2}?
$${If}\:{the}\:{density}\:{function}\:{of}\:{X}\:{equals} \\ $$$${f}\left({x}\right)\:=\:\begin{cases}{{c}\:{e}^{−\mathrm{2}{x}} \:,\:\mathrm{0}<{x}<\infty}\\{\mathrm{0}\:,\:\:\:\:\:\:\:\:\:\:{x}<\mathrm{0}}\end{cases} \\ $$$${find}\:{c}.\:{What}\:{is}\:{P}\left\{{X}>\mathrm{2}\right\}? \\ $$
Commented by prakash jain last updated on 05/Jul/16
∫_0 ^( ∞) ce^(−2) dx≠1  So f(x) is not a valid probability density  function.
$$\int_{\mathrm{0}} ^{\:\infty} {ce}^{−\mathrm{2}} {dx}\neq\mathrm{1} \\ $$$$\mathrm{So}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{valid}\:\mathrm{probability}\:\mathrm{density} \\ $$$$\mathrm{function}.\: \\ $$
Commented by WAI LIN last updated on 05/Jul/16
Now I have answer.
$${Now}\:{I}\:{have}\:{answer}. \\ $$
Commented by WAI LIN last updated on 05/Jul/16
Sorry! Befroe my question is wrong.
$${Sorry}!\:{Befroe}\:{my}\:{question}\:{is}\:{wrong}. \\ $$
Answered by WAI LIN last updated on 05/Jul/16
∫_0 ^∞  f(x)dx =1  ∫_0 ^∞  c e^(−2x) dx = 1  −(c/2) [e^(−2) ]_0 ^∞  =1  (c/2) =1  ∴c = 2  Next,P(X>2)=∫_2 ^∞  2 e^(−2x)  dx = −[e^(−2x) ]_2 ^∞  = e^(−4)
$$\int_{\mathrm{0}} ^{\infty} \:{f}\left({x}\right){dx}\:=\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{c}\:{e}^{−\mathrm{2}{x}} {dx}\:=\:\mathrm{1} \\ $$$$−\frac{{c}}{\mathrm{2}}\:\left[{e}^{−\mathrm{2}} \right]_{\mathrm{0}} ^{\infty} \:=\mathrm{1} \\ $$$$\frac{{c}}{\mathrm{2}}\:=\mathrm{1} \\ $$$$\therefore{c}\:=\:\mathrm{2} \\ $$$${Next},{P}\left({X}>\mathrm{2}\right)=\int_{\mathrm{2}} ^{\infty} \:\mathrm{2}\:{e}^{−\mathrm{2}{x}} \:{dx}\:=\:−\left[{e}^{−\mathrm{2}{x}} \right]_{\mathrm{2}} ^{\infty} \:=\:{e}^{−\mathrm{4}} \\ $$

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