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I got this question: We consider a sample of $Z_1,...,Z_n$ of independent and identically distributed random variables where $Z_i=X_i-Y_i$ with $X_i$ and $Y_i$ independent and exponentially distributed with mean $\theta$

Then I have to show that $$E_{\theta}(Z_i^4)=24 \theta^4$$

Can someone help med with that? I think I can use that $$V(Z)=E(Z^2)-E(Z)^2$$ and $$V(Z^2)=E(Z^4)-E(Z^2)^2$$ But how can I find $E(Z)$ and $V(Z)$ and $V(Z^2)$?

Lifeni
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  • one way to solve this could be finding pdf of Z, then finding mgf of Z and taking 4th moment at t=0. Note that n-th moment of moment generation function ($E(e^{tX})$) at t=0 gives you $E(Z^n)$ – Snowball Apr 27 '21 at 10:33
  • Finding PDF by transformations sentence? How can I find MGF of Z and taking 4th moment? – Lifeni Apr 27 '21 at 10:40
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    I posted my answer. It is a bit harder than other answers, but more useful for general cases. Using that, you can calculate all $E[Z^n]$. Note that I found mgf for only $|t| < \lambda$ because we are going to evaluate only at $t=0$. – Snowball Apr 27 '21 at 11:19

3 Answers3

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I think it is easy in this way:

First of all observe that

$$\mathbb{E}[X^k]=\int_0^{\infty} x^k\frac{1}{\theta}e^{-x/\theta}dx=\theta^k\underbrace{\int_0^{\infty}\left(\frac{ x}{\theta}\right)^ke^{-x/\theta}d\left(\frac{x}{\theta}\right)}_{=\Gamma(k+1)=k!}=k!\theta^k$$

Second, expand $(X-Y)^4$ and find the result using independence property

thus

$$\mathbb{E}[X-Y]^4=\theta^4[4!-4\cdot3!+6\cdot2!\cdot2!-4\cdot3!+4!]=24\theta^4$$

...the result you have to prove (after your editing) is done!

tommik
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$X_j,Y_j \sim \exp(\lambda)$, independent. So $$\mathbb{E}[Z^4]=\mathbb{E}[(X_j^2+Y_j^2-2X_jY_j)^2]=\mathbb{E}[X_j^4+X_j^2Y_j^2-2X^3_jY_j+X_j^2Y_j^2+Y_j^4-2X_jY_j^3-2X_j^3Y_j-2X_jY_j^3+4X_j^2Y_j^2]=$$ $$=2\frac{4!}{\lambda^4}+6\frac{2!2!}{\lambda^4}-8\frac{3!}{\lambda^4}=\frac{24}{\lambda^4}$$

(Mean $\theta$ implies $\theta=\frac{1}{\lambda}$)

Snoop
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You can find out pdf of Z here: Pdf of the difference of two exponentially distributed random variables

Note that mean $\theta$ means parameter is $1/\theta = \lambda$. Replacing variables with $\lambda$, we get $f_Z(z) = \frac{\lambda}{2}e^{-\lambda |z|}$. So mgf of Z is $$M_Z(t) = E[e^{tZ}] = \int_{-\infty}^{\infty} e^{tz}\frac{\lambda}{2}e^{-\lambda |z|} \,dz = \frac{\lambda}{2 (t - \lambda)}e^{z(t - \lambda)} \Big|_{0}^\infty + \frac{\lambda}{2 (t + \lambda)}e^{z(t + \lambda)} \Big|_{-\infty}^0$$

For $t < \lambda$ ($\lambda > 0$ and we need result at $t=0$), $\lim_{z\to\infty} e^{z(t - \lambda)} = 0$ and $\lim_{z\to{-\infty}} e^{z(t + \lambda)} = 0$. Therefore, $M_Z(t) = E[e^{tZ}] = \frac{\lambda}{2}(-\frac{1}{t-\lambda} + \frac{1}{t+\lambda})$. Then take derivative w.r.t $t$ for four times and evaluate at $t=0$

$$E[Z^4] = M_Z^{(4)}(0) = \frac{\lambda}{2}(\frac{-24}{-\lambda^5} + \frac{24}{\lambda^5}) = 24\lambda^{-4} = 24\theta^{4}$$

You can also generalize from here: $$E[Z^n] = M_Z^{(n)}(0) = n!\lambda^{-n} = n!\theta^{n}$$

Snowball
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