2
$\begingroup$

$$X \sim \mathcal{N}(\mu_X,\,\sigma_X^2)$$ $$Y \sim \mathcal{N}(\mu_Y,\,\sigma_Y^2)$$ $\mu_X$ and $\sigma_X$ have unit decibel watt ($\text{dBW}$); $\mu_Y$ and $\sigma_Y$ have unit watt ($\text{W}$).

$$Z = X + Y$$ What is the PDF of $Z$?

I'm not sure if it's possible to find it, or if it even makes sense.

I would expect it to produce a non-normal, lopsided curve, when the PDF of $Z$ is plotted with input in $\text{dBW}$.



To express an arbitrary power $P$ in watts as $x$ in $\text{dBW}$, or vice versa, the following expressions may be used: $$x = 10 \log_{10} \frac{P}{ 1 \mathrm{W}}$$ and $$P = 1 \text{W} \cdot 10^{\frac{x}{10}}$$ where $P$ is the power in $\text{W}$ and $x$ is the power level in $\text{dBW}$.

$\endgroup$
9
  • $\begingroup$ $X$ and $Y$ independent? $\endgroup$ Jul 20, 2014 at 17:51
  • $\begingroup$ @SeyhmusGüngören Yes $\endgroup$
    – Max
    Jul 20, 2014 at 18:07
  • $\begingroup$ If $X\sim N\left(\mu_{X},\sigma_{X}^{2}\right)$ and $Y\sim N\left(\mu_{Y},\sigma_{Y}^{2}\right)$ are independent then $Z=X+Y\sim N\left(\mu_{X}+\mu_{Y},\sigma_{X}^{2}+\sigma_{Y}^{2}\right)$. Unfortunately I don't know about things like '... have unit $A$' or '..$a$ in $A$' and I wouldn't be surprised if I am not the only one in this. Make your question more accessible by some explaining. $\endgroup$
    – drhab
    Jul 20, 2014 at 18:14
  • $\begingroup$ the transformation is ugly. do not expect $Z$ to be standard $\endgroup$
    – Lost1
    Jul 20, 2014 at 18:14
  • 2
    $\begingroup$ $X$ transformed into Watts has a log-normal distribution. So assuming that the addition is in Watts, then $Z$ will not have a simple distribution. Meanwhile, if $Y$ had a normal distribution then it would have a positive probability of being negative and so could not be transformed into a dBW. distribution. $\endgroup$
    – Henry
    Jul 20, 2014 at 21:21

2 Answers 2

3
$\begingroup$

We are given $Z$ is the sum of 2 independent normals so $Z$ will be normal. We just have to straighten out the means and variances. Evidently we can measure power in $W$ or dBW. Never heard of dBW but it is just $W$ measured on a log scale. (Like measuring strength of earthquakes on the Richter scale.) So we have

$$\text{dBW}=10\log_{10}W.$$

We can discuss 4 distributions in this problem: $X$ measured in $W$ and in dBW; $Y$ measured in $W$ and in dBW.

Notice the question does not tell us the units of $X$ and $Y$, only the units in which the mean and std dev are given. Just like $X$ could be a distribution of heights in feet but I will state the mean and sd in meters. (That is a lot simpler since it is a linear transformation of units.)

However, we must have $W\gt 0$ while dBW can be negative. So now we notice $X$ and $Y$ are given as normal implies that those are distributions in dBW. Since $\mu_x$ and $\sigma_x$ are also stated in dBW units that is consistent with rv $X.$ i.e., $E(X)=\mu_x$ and $SD(X)=\sigma_x.$ So we have everything figured out for $X.$ For the $Y$ distribution, let $W$ be the equivalent rv measured in watts. Therefore, $W$ is defined by: $$ Y=10\log_{10}W$$ $$W=10^{Y/10} $$ $$ W=e^{cY}, \text{where }c:=(\ln 10)/10.$$

There are 2 possible solutions to the problem depending upon what was meant by the given information. Note that in both solutions, I have to "interpret" what is meant by the parameters in $Y\sim N(\mu_y,\sigma_y^2).$

Solution $1$:

The mean and std dev of $Y$ are given to us in Watts and we need to convert those two numbers to dBW. OK, that's easy: $$ m:= 10\log_{10} \mu_y \text{ and } s:= 10\log_{10} \sigma_y$$ Now $Z$ in dBW is normal with mean $\mu_x + m$ and variance $\sigma^2_x+s^2.$

That's almost too easy so now here's another interpretation.

Solution $2$:

Now I will assume we are being given the mean and std dev of the rv $W$. So I assume the statement really meant: $\mu_y,\sigma_y$ are given values which are defined as $\mu_y=E(W), \sigma_y^2=Var(W).$ Now find $m,s^2$ where $Y\sim N(m,s^2).$

Now we want to find the relationship between the mean and variance of $W$ and the parameters of $Y$. Since $Y$ is given as normal, finding the mean of a lognormal rv is a well-known result; we just have an extra factor of $c:$ $$ W=e^{cY} $$ $$ E(W)=\mu_y=\exp(cm+c^2s^2/2)$$ $$ Var(W)=\sigma_y^2=[\exp(c^2s^2)-1]\exp[2cm+c^2s^2]$$

Now solve the two equations for $m$ and $s^2.$

$$m=\frac{1}{c}\ln\left[\frac{\mu^2_y}{\sqrt{\mu_y^2+\sigma_y^2}} \right] $$ $$s^2=\frac{1}{c^2}\ln\left[\frac{\mu_y^2+\sigma_y^2}{\mu_y^2} \right]$$

and therefore, $Z$ is normal with parameters: $$E(Z)=\mu_x+m, Var(Z)=\sigma^2_x+s^2$$

... and we're done.

$\endgroup$
0
$\begingroup$

First of all, if $X$ and $Y$ are independent and Gaussian, then their sum is also Gaussian. There is no doubt about it. You question is a bit different. You will have a transformation to one of your random variables before the convolution. This means first, both of the domains of the PDFs should be the same. When you take Log of a random variable distributed as $X$ you will get something like this, call it as $X^{'}$. Then since $X$ and $Y$ are independent, so $X^{'}$ and $Y$, which means that what remains is to convolve the density of $X^{'}$ with the density of $Y$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .