How does a pdf of the difference of two random variables relate to the pdf of each random variable

Question

Let $T_1$ and $T_2$ be non-negative continuous random variables (rv) denoted in the form $T_i = \mu_i + \sigma_i X_i$ for $i=1,2$ where \begin{eqnarray*} T_{1} &=&\mu _{1}+\sigma _{1}X_{1} \\ T_{2} &=&\mu _{2}+\sigma _{2}X_{2} \end{eqnarray*}

Suppose that $X_1$ and $X_2$ are iid with $f_{X_i} (x)=\lambda e^{-\lambda x}$ for $x\geq 0$ or zero otherwise.

Now define $Z=T_2 - T_1$ such that $Z=\mu_z + \sigma_z Y$; where $Y$ is a standardized rv with pdf $f_{Z}(z)$.

Question:

Express $f_Y(y)$ in terms of $f_{X_1}(x) \text{ and/or } f_{X_2}(x)$.

Trial

Following steps suggested by sds; I arrived at the following.

I expressed the pdf of $T_{i}$ in terms of the pdf of $X_i$ starting from the CDF of these variables as follows.

\begin{equation*} F_{T_{i}}\left( x\right) =F_{X_{i}}\left( \frac{x-\mu _{i}}{\sigma _{i}} \right) \mbox{ and } F_{-T_{i}}\left( x\right) = 1-F_{X_{i}}\left( \frac{-x-\mu _{i}}{\sigma _{i}}\right) \end{equation*} hence \begin{equation*} f_{T_{i}}\left( x\right) =\frac{1}{\sigma _{i}}f_{X_{i}}\left( \frac{x-\mu _{i}}{\sigma _{i}}\right) \text{ and }f_{-T_{i}}\left( x\right) =\frac{1}{% \sigma _{i}}f_{X_{i}}\left( \frac{-x-\mu _{i}}{\sigma _{i}}\right) \end{equation*}

Defining $-T_1 = Z-T_2 $ and using the convolution of pdfs, we have: \begin{align*} f_{Z}\left( z\right) &=\int_{-\infty }^{\infty }f_{T_{2}}\left( x\right) f_{-T_{1}}\left( z-x\right) dx \\ &=\int_{-\infty }^{\infty }\frac{1}{\sigma _{2}}f_{X_{2}}\left( \frac{x-\mu _{2}}{\sigma _{2}}\right) \frac{1}{\sigma _{1}}f_{X_{1}}\left( z-\left( \frac{-x-\mu _{2}}{\sigma _{2}}\right) \right) dx \\ &=\left( \sigma _{1}\sigma _{2}\right) ^{-1}\int_{-\infty }^{\infty }f_{X_{2}}\left( \frac{x-\mu _{2}}{\sigma _{2}}\right) f_{X_{1}}\left( z+% \frac{x+\mu _{2}}{\sigma _{2}}\right) dx \end{align*}

The integral is nonzero for $ \left( x>-\sigma _{2} z-\mu _{2}\right) \text{ and }\left( x>\mu _{2} \right) $ which is:

\begin{equation*} \max \left( -\sigma _{2}z-\mu _{2},\mu _{2}\right) =\left\{ \begin{array}{l} -\sigma _{2}z-\mu _{2}\text{ if }z\leq \frac{-2\mu _{2}}{\sigma _{2}} \\ \mu _{2}\text{ otherwise}% \end{array}% \right. \end{equation*}

First, I attempted deriving $f_{Z}\left( x\right) $ for $z\leq \frac{-2\mu _{2}% }{\sigma _{2}}$ as:

\begin{equation*} f_{Z}\left( z\right) =\left( \sigma _{1}\sigma _{2}\right) ^{-1}\int_{-\sigma _{2}z-\mu _{2}}^{\infty }f_{X_{2}}\left( \frac{x-\mu _{2}% }{\sigma _{2}}\right) f_{X_{1}}\left( z+\frac{x+\mu _{2}}{\sigma _{2}}% \right) dx \end{equation*}

However, I am not sure about the limits of integration and if everything is correct, after all. I will appreciate any help on this!

Answer 1

You need to take 3 steps:

1. Express the pdf of $T_1=\mu_1+\sigma_1 X_1$ in terms of $f_{X_1}$.
2. Express the pdf of $-T_2=-\mu_2-\sigma_2 X_2$ in terms of $f_{X_2}$.
3. Use the fact that the pdf of a sum of independent RVs is the convolution of the pdfs of the summands.