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I am trying to find the following: If $X\sim\mathrm{Exp}(1)$ and $Y\sim\mathrm{Exp}(2)$, what is the pdf of $Z=X+Y$?

I tried to use the convolution formula but am not sure what the limits of the integral are:

$$g(z) = \int_{0}^{z} 1 \cdot 2\cdot\exp(-1\cdot(y-x))\cdot\exp(-2x) \mathrm dx.$$

I ended up getting:

$g(z) = 2\cdot(1-e^{-z})\cdot e^{-y}$.

Would this be correct? Thanks!

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    $\begingroup$ No that's not correct. You might want to review how to set up convolution integrals - not just the limits, but also the integrand. $\endgroup$ – Glen_b Feb 20 '14 at 8:29
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\begin{equation} g(z) = 2 \int_{- \infty}^{+ \infty} e^{-(z-x)} e^{-2x} \mathrm{d}x \\ =2 \int_{0}^{z} e^{-(z-x)} e^{-2x} \mathrm{d}x \\ = 2 e^{-z} \int_{0}^{z} e^{-x} \mathrm{d}x \\ = 2e^{-z} (1 - e^{-z}), \end{equation}

Note that this result holds if $z>0$. In general, when $p(x) = \alpha e^{-\alpha x}$ and $p(y) = \beta e^{-\beta y}$, the convolution of these two is equal to:

\begin{equation} g(z) = \frac{\alpha \beta}{(\alpha - \beta)} (e^{-\beta z} - e^{-\alpha z}) \end{equation}

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First, are $X$ and $Y$ independent? (Suppose they are, since you're using the convolution formula.)

It should be easier to figure out the limit if you start with the bivariate transformation.

Let $W=Y$. For the transformation from $(X,Y)$ to $(Z,W)$, the Jacobian is 1, and

$f_{Z,W}(z,w)=f_{X,Y}(z-w,w)|J|=f_{X,Y}(z-w,w)=f_X(z-w)f_Y(w)$.

Then, to obtain the desired pdf $g(z)$ simply integrate out $w$.

Note that $w=y=z-x \le z$, so the limit of integral is $(0,z)$.

$g(z)=\int_{0}^{z} f_{Z,W}(z,w)dw=\int_{0}^{z} f_X(z-w)f_Y(w)dw$, which is the convolution formula.

Finally,

$g(z)=2\int_{0}^{z} e^{-z+w}e^{-2w}dw=2e^{-z}\int_{0}^{z} e^{-w}dw=2e^{-z}(1-e^{-z})$.

(only for $z\ge0$. If $z<0$, $g(z)=0$.)

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