Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have this confusion related to calculating the probability distribution of a variable. If I have a variable $x_1$ which has a pdf $p(x_1)$.Lets assume that the distribution is gaussian with mean $X_1$. I sample a point from this distribution lets say $X_1'$.

Now I have another random variable $x_2$ which has a distribution $p(x_2)$. Its mean is equal to a constant, $X_2$ plus the previous sample $X_1'$. So how can I write the pdf function for this random variable $x_2$. I mean I want its mean to be dependent upon the sample from the distribution of $x_1$.

share|cite|improve this question
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

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 $\mu_1+\mu_X$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.