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$ f(x) = \begin{cases} \frac15 e^{(-\frac15x)}, x>0 \\ 0, \text{elsewhere}\\ \end{cases}$

How to calculate $E[(X+5)]$ and $E[(X+5)^2]$ ?

Thanks a lot.

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    Do you know what $E[X]$ is?2012-10-22
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    $$\displaystyle \int_0^{\infty} \dfrac{(x+5)\exp(\frac{-x}{5})}{5}$$2012-10-22
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    What did you try and what is STOPPING you?2012-10-22
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    I know $E[X]$ means the mean/expected value of $f(x)$2012-10-22
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    I know $E[X]$ = 5 but I've no idea about $E[(X+5)^2]$2012-10-22

2 Answers 2

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Hints: for $E[X+5]$ use the linearity of expectation. What is $E[5]$?

For $E[(X+5)^2]$ you can go back to the definition $$E[(X+5)^2]=\displaystyle \int_0^{\infty} \dfrac{(x+5)^2\exp(\frac{-x}{5})\; dx}{5}$$ or you can expand out $E[(X+5)^2]=E[X^2]+2E[X]E[5]+E[5^2]$

  • 0
    Is $E[X] = \frac1\lambda $? Then $E[X] = 5$. Can we use the similar method to find out $E[(x+5)^2]$ ?2012-10-22
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    Yes, $E[X]=5$ The way you find that (if you don't look it up) is the integral Inquest gave without the +5. The integral in my answer is the same method for $E[(X+5)^2]$ Can you do it?2012-10-22
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    Sorry. Can you show the steps why $E[(X+5)^2]=E[X^2]+2E[X]+E[5^2]$?2012-10-23
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    Why it's not $E[(X+5)^2]=E[X^2]+10E[X]+E[5^2]$ but $+2E[X]$?2012-10-23
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    @Rick: You are right, I dropped the $E[5]$. Fixed.2012-10-23
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If you know that $$ \int_0^\infty u^2 e^{-u} \, du = 2, $$ then you can write $$ \int_0^\infty x^2 e^{-x/5}\, \frac{dx}{5} = 5^2 \int_0^\infty \left(\frac x 5\right)^2 e^{-x/5} \, \frac{dx}{5} = 5^2\int_0^\infty u^2 e^{-u} \, du = 5^2\cdot 2. $$ That gets you $E(X^2)$. And $E((X+5)^2) = E(X^2) + 10E(X) + 5^2$, etc.

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    Can you explain a little bit more on $E((X+5)2)=E(X2)+10E(X)+5$? I'm quite confused.2012-10-22
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    Why it's not $E[(X+5)^2]=E[X^2]+10E[X]+E[5^2]$ but $+5$?2012-10-23
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    Sorry --- typo: It's $5^2$.2012-10-23