Let $Z=X+Y$; where $X\sim \mathscr N(0,\sigma^2_1)$ i.e. a Gaussian random variable and $Y$ follows the Rayleigh distribution: $$ f_Y(y) = \frac{y}{\sigma^2_2}\exp\left(-\frac{y^2}{2\sigma^2_2}\right) \mathbf{1}_{y \geqslant 0} $$ What will be the distribution of $Z$?
Asked
Active
Viewed 3,586 times
0
-
Would you like to know if the distribution of $x+y$ belongs to a known class of distributions? Also: are $x,y$ independent? – SBF May 09 '12 at 13:19
2 Answers
2
The probability density function for $Z$, $f_Z(z)$ is obtained as a convolution of $f_X$ and $f_Y$, assuming $X$ and $Y$ are independent: $$ f_Z(z) = \int_{0}^\infty f_Y(y) f_X(z-y) \mathrm{d} y = \frac{1}{\sqrt{2 \pi} \sigma_1 \sigma_2^2} \int_0^\infty y \exp\left( -\frac{y^2}{2 \sigma_2^2}\right) \exp\left( - \frac{(z-y)^2}{2 \sigma_1^2} \right) \mathrm{d} y $$ Evaluation of this integral is straightforward, but tedious, and is done via integration by parts, with the following result: $$ f_Z(z) = \frac{\sigma_2 z}{\left(\sigma_1^2+\sigma_2^2\right)^{3/2}} \mathrm{e}^{-\frac{z^2}{2 \left(\sigma_1^2+\sigma_2^2\right)}} \Phi\left( \frac{\sigma_2}{\sigma_1} \frac{ z}{ \sqrt{\sigma_1^2+\sigma_2^2}}\right) + \frac{\sigma_1}{\sqrt{2 \pi} \left(\sigma_1^2+\sigma_2^2\right)} \mathrm{e}^{-\frac{z^2}{2 \sigma_1^2}} $$ where $\Phi(x)$ is the cumulative distribution function of the standard normal random variable.
Sasha
- 70,631
-
Any reference please? (Assuming that you didn't need to do it by hand) – Loves Probability Nov 23 '16 at 15:05
-
3I did it manually or with Mathematica. Steps are: combine two exponents and complete the square, then integrate by parts. It is very straightforward, but tedious. – Sasha Nov 23 '16 at 16:32
-
1Yup, thanks! I was expecting a magical tool from you other than your brain. It seems there is nothing like it. (ICYDK, Thats just a bit of appreciation) :) – Loves Probability Nov 24 '16 at 01:41
-
@Sasha -- A detailed step by step solution would be greatly appreciated, as this question arises all the time in wireless communications. It is one of the most fundamental distributions in wireless communications, and being able to derive this is incredibly important. – The Dude Apr 24 '19 at 13:50
-
I just found an application for mechanical as well. This is very helpful. – Mad Physicist Apr 09 '22 at 16:10
-1
The correct answer is
$$\frac{\exp\left(-\frac{|x|}{\sigma_w \sigma_h} \right)}{2 \sigma_w \sigma_h}$$
-
-
2Please use LaTeX for math. And don't comment on your own answers to say they are right. – vonbrand Sep 30 '15 at 18:13