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Let $X_1$ and $X_2$ be independent and identically distributed continuous random variables, with probability density function

$$p(x)=\begin{cases} \exp(-x), & \text{if}\ x>0 \\ 0, & \text{otherwise}. \end{cases}$$

Let

$$Z=\frac{X_1}{X_1+X_2}+2X_2$$

Derive the probably density function of $Z$. It is sufficient to give the required pdf in the form of an integral of a joint pdf.

I have found the pdfs of $Y_1=\frac{X_1}{X_1+X_2}$ and $Y_2=2X_2$,

$$g(y_1)=\begin{cases} 1, & \text{if}\ 0<y_1<1 \\ 0, & \text{otherwise} \end{cases}$$

and

$$g(y_2)=\begin{cases} \frac12\exp\left(-\frac{y_2}{2}\right), & \text{if}\ y_2>0 \\ 0, & \text{otherwise}. \end{cases}$$

But now I cannot see how to find the transformation of $Z$ as $Y_1$ and $Y_2$ do not have a joint pdf and are not necessarily independent.

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It's OK. Z is the sum and then the density is the convolution.

Exist this theorem: if $X_i$ are independent and $f_i$ borel-measuring functions, then $f_i(X_i)$ are independent

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  • $\begingroup$ I.e.: yes, are independent $\endgroup$ – Martín Vacas Vignolo Dec 10 '15 at 4:27
  • $\begingroup$ But $Y_1$ and $Y_2$ both depends on $X_2$, so your argument fails here. $\endgroup$ – Stefan Hansen Dec 10 '15 at 6:48
  • $\begingroup$ The transformations of $Y_1$ and $Y_2$ both depend on $X_2$, but the pdfs don't. $\endgroup$ – Will Dec 10 '15 at 9:12
  • $\begingroup$ You can use the change variables theorem $\endgroup$ – Martín Vacas Vignolo Dec 10 '15 at 17:58

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