How to calculate MEAN of exponential distribution?

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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 $E[(X+5)^2]=E[X^2]+2E[X]E[5]+E[5^2]$

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Rick

Updated on June 16, 2022

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

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

Thanks a lot.

• Nikita Evseev over 9 years
Do you know what $E[X]$ is?
• Inquest over 9 years
$$\displaystyle \int_0^{\infty} \dfrac{(x+5)\exp(\frac{-x}{5})}{5}$$
• Did over 9 years
What did you try and what is STOPPING you?
• Rick over 9 years
I know $E[X]$ means the mean/expected value of $f(x)$
• Rick over 9 years
I know $E[X]$ = 5 but I've no idea about $E[(X+5)^2]$
• Rick over 9 years
Is $E[X] = \frac1\lambda$? Then $E[X] = 5$. Can we use the similar method to find out $E[(x+5)^2]$ ?
• Ross Millikan over 9 years
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?
• Rick over 9 years
Can you explain a little bit more on $E((X+5)2)=E(X2)+10E(X)+5$? I'm quite confused.
• Rick over 9 years
Sorry. Can you show the steps why $E[(X+5)^2]=E[X^2]+2E[X]+E[5^2]$?
• Rick over 9 years
Why it's not $E[(X+5)^2]=E[X^2]+10E[X]+E[5^2]$ but $+2E[X]$?
• Rick over 9 years
Why it's not $E[(X+5)^2]=E[X^2]+10E[X]+E[5^2]$ but $+5$?
• Ross Millikan over 9 years
@Rick: You are right, I dropped the $E[5]$. Fixed.
• Michael Hardy over 9 years
Sorry --- typo: It's $5^2$.