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The lifetime in hours of a tubelight is a random variable $X$ having a probability density function $f(x) = 4xe^{-2x}$, where $x>0$. If $Y= -3X+10$, then find the Expected Value of $Y$, and variance of $Y$.

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If its a homework problem, please add the tag homework to it –  Soham Sep 9 '12 at 5:53
    
My comment on your other question fully applies. –  Did Sep 9 '12 at 7:52

2 Answers 2

We have $E(Y)=-3E(X)+10$ and $\text{Var}(Y)=(-3)^2\text{Var}(X)=9\text{Var}(X)$. For in general if $Y=aX+b$, then $E(Y)=aE(X)+b$ and $\text{Var}(Y)=a^2\text{Var}(X)$.

So we need the mean and variance of $X$. These are straight integration problems. For the mean of $X$, we want $$\int_0^\infty 4x^2e^{-2x}\,dx.$$ Use Integration by Parts, letting $u=4x^2$ and $dv=e^{-2x}\,dx$. Thus $du=8x\,dx$ and $v$ can be taken to be $-\frac{1}{2}e^{-2x}$. So our integral is $$\left. -2x^2e^{-2x}\right|_0^\infty+ \int_0^\infty 4xe^{-2x}\,dx.$$ The first part dies at both $0$ and $\infty$. The remaining integral is $1$, since it is the integral of our density function.

For the variance of $X$, we calculate $E(X^2)-(E(X))^2$. Another integration by parts. When the smoke clears I think you will find that $X$ has variance $\dfrac{1}{2}$.

Now that we know the mean and the variance of $X$, the mean and variance of $Y$ are easy to compute, as mentioned at the beginning of this answer.

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We have $Y=-3X+10$

Hence $E[Y]=E[-3X+10]$

$\Rightarrow E[Y] = -3E[X]+10 $ (since expectation is a linear operator)

We have

$ E[X] = \int_{0}^{\infty } x \cdot 4xe^{-2x}dx$

$\Rightarrow \int_{0}^{\infty}4x^2 e^{-2x}dx$

We notice that, that this is equivalent to Laplace transform of $x^2$ with $s=2$

We know Laplace Transform of the $\int_{0}^{\infty}t^n e^{-sn}dt$ $\Rightarrow n!/s^{n+1}$

Thus $I=\Rightarrow \int_{0}^{\infty}4x^2 e^{-2x}dx= 2!/s^3$

Putting $s=2$ we get $I=1/4$ $\Rightarrow E[X]=1$ $\Rightarrow E[Y]=7$

We can in a similar way find $E[Y^2]$ and thus $Var[Y]$

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