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I have this confusion related to calculating the probability distribution of a variable. If I have a variable x1 which has a pdf p(x1).Lets assume that the distribution is gaussian with mean X1. I sample a point from this distribution lets say X1'.

Now I have another random variable x2 which has a distribution p(x2). Its mean is equal to a constant,X2 + the previous sample X1'. So how can I write the pdf function for this random variable x2. I mean I want its mean to be dependent upon the sample from the distribution of x1.

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Your choice of notation is dreadful. Most people tend to use upper-case (capital) letters for random variables and lower-case letters for the arguments of density functions, real variables, etc. Also, is p the same function in the two cases? –  Dilip Sarwate Mar 31 '13 at 3:24

1 Answer 1

If $X$ is a continuous random variable with pdf $f_X(x)$ with mean $\mu_X$, and $Y$ is a random variable whose conditional pdf given that $X = a$ is $f_{Y\mid X=a}(y\mid X=a)$ with mean $\mu_1+a$, that is, $E[Y\mid X=a] = \mu_1+a$, then the random variable $E[Y\mid X] = \mu_1 + X$, and $$E[Y] = E[E[Y\mid X]] = E[\mu_1+X] = \mu_1+\mu_X.$$ The unconditional density of $Y$ is $$f_Y(y) = \int_{-\infty}^\infty f_{Y\mid X=a}(y\mid X=a)f_X(a)\,\mathrm da.$$ If $X$ is a unit-variance Gaussian random variable, and the conditional pdf of $Y$ is also a unit-variance Gaussian density, then $$f_Y(y) = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}(y-\mu_1-a)^2\right) \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}(a-\mu_X)^2\right)\,\mathrm da$$ which, after you combine the exponentials, complete the square, etc. will work out to be a Gaussian density with mean $\mu_1+\mu_X$.

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