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So let suppose I have a random variable Y that is defined as follows:

$$Y=\alpha x+ \mathcal{N}(\mu,\sigma) \text{ where } x \text{ } \epsilon \text{ }\mathbb {R}$$

and

$$\mathcal{N}(\mu,\sigma)\text{ is a i.i.d. normally distributed random variable with mean }\mu\text{ and variance }\sigma$$

So I know Y is a random variable, but x is not, however is seems to me that their is a probability measure

$$P(x,Y)\text{ } \epsilon\text{ }[0,1]$$

Therefore, the mutual information is

$$I(x;Y)=H(x)+H(x|Y)=H(Y)+H(Y|x)$$

However it seems that

$$H(x)=-\int_{x}p(x)lnp(x)dx$$

is not defined because x is not a random variable so is their really any mutual information between x and Y? is p(x,Y) an actual joint probability distribution? Any insight would be awesome thanks.

$$Y=\alpha x+ \mathcal{N}(\mu,\sigma) \text{ where } x \text{ } \epsilon \text{ }\mathbb {R}$$

and

$$\mathcal{N}(\mu,\sigma)\text{ is a i.i.d. normally distributed random variable with mean }\mu\text{ and variance }\sigma$$

So I know Y is a random variable, but x is not, however is seems to me that their is a probability measure

$$P(x,Y)\text{ } \epsilon\text{ }[0,1]$$

Therefore, the mutual information is

$$I(x;Y)=H(x)+H(x|Y)=H(Y)+H(Y|x)$$

However it seems that

$$H(x)=-\int_{x}p(x)lnp(x)dx$$

is not defined because x is not a random variable so is their really any mutual information between x and Y? is p(x,Y) an actual joint probability distribution? Any insight would be awesome thanks.

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