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Suppose you would like to measure a signal, which can be modelled as a Gaussian random variable with $1.0V$ mean and $0.2V$ standard deviation. If you measure this signal with your instrument, it will be subject to the instrument noise and the measured signal will be sum of the noise and the signal. Noise of your instrument can also be modelled as a Gaussian random variable with $0.1V$ mean and $0.1V$ standard deviation.

Write out exact expression to represent the pdf of the measured signal $Z$

My attempt:

Let $X$ represent the random variable for signal & $Y$ represent the random variable for noise then the measured signal $Z$ is given by:

$Z = X+Y$

To find PDF of $Z$, we first find $CDF$ and then take the derivative of it to get the PDF

$CDF$ is given by:

$F_Z(z)=F_Z(Z\leq z)=F_Z(X+Y\leq z)$

I don't know how to go further. Any help and guidance will be appreciated

Ameer786
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  • Are $X$ and $Y$ independent? If so then $Z=X+Y$ has normal distribution. – drhab Apr 23 '21 at 10:12
  • @drhab it is not mentioned in the question whether $X$ and $Y$ are independent – Ameer786 Apr 23 '21 at 10:13
  • Probably they are meant to be independent. If not then you need the joint distribution of $(X,Y)$. That has not been given to you (which makes you powerless). If that is a normal distribution then again $X+Y$ has normal distribution, but you need covariance of $X,Y$ to find the parameters of this distribution. – drhab Apr 23 '21 at 10:15

2 Answers2

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You want "the sum of two independent Gaussian random variables is also a Gaussian random variable". This is a well-known result, and it's explained brilliantly on this answer. You will have to adapt the argument because your variables don't have mean zero, but this should be a minor detail.

R_B
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If you assume that signal $X$ and noise $Y$ are independent, then the characteristic function of $Z=X+Y$ is $$\phi_Z(\omega)=\mathbb{E}[e^{i\omega Z}]$$ from which we can compute $$\phi_Z(\omega)=\mathbb{E}[e^{i\omega (X+Y)}]\underbrace{=}_{\textrm{indep.}}\mathbb{E}[e^{i\omega X}]\mathbb{E}[e^{i\omega Y}]=\phi_X(\omega)\phi_Y(\omega)$$ it is known that the characteristic function of a Gaussian density is $$\phi_j(\omega)=e^{i\mu_j-0.5\omega^2\sigma^2_j}$$ So if $X \sim N(\mu_X,\sigma_X^2)$ and $Y \sim N(\mu_Y,\sigma_Y^2)$ $$\phi_Z(\omega)=e^{i(\mu_X+\mu_Y)-0.5\omega^2(\sigma^2_X+\sigma^2_Y)}$$ which implies $Z \sim N(\mu_X+\mu_Y,\sigma^2_X+\sigma^2_Y)$. So the pdf of $Z=X+Y$ is $$f_Z(z)=\frac{1}{\sqrt{2\pi(\sigma^2_X+\sigma^2_Y)}}e^{-\frac{(z-(\mu_X+\mu_Y))^2}{2(\sigma^2_X+\sigma^2_Y)}}$$

Snoop
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