Contents
Practice Problem: obtaining the joint pdf from the marginals of two independent variables
A random variable X has the following probability density function:
$ f_X (x) = \frac{1}{\sqrt{2\pi}} e^{\frac{-x^2}{2}}. $
Another random variable Y has the following probability density function:
$ f_Y (y) = \frac{1}{3 \sqrt{2\pi} } e^{\frac{-(x-7)^2}{6}}. $
Assuming that X and Y are independent, find the joint probability function fXY(x,y).
- - Is this a valid pdf (coeff divided by 3 instead of sqrt(3)), or is it unnecessary to satisfy prob. axioms here?
- Good point. That's a typo. I'm glad you pointed it out. Can somebody please find the correct normalization? -pm
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
Because X and Y are independent, the joint probability function can be represented as the product of the two marginal density functions:
$ f_{XY}(x,y) = f_X(x)f_Y(y) $
Thus, the joint probability function is simply the two marginal density functions multiplied together:
$ f_{XY}(x,y) = \frac{1}{6\pi} e^{\frac{1}{6}(-4x^2+14x-49)}. $
- Instructor's comment: Do you know why the joint pdf is the product of the marginals? All we have seen in class is that the probability of two independent events is the product of the probabilities of the respective events. As you know, the pdf is not a probability. -pm
- Is this because the pdf is the envelope function of the probabilities just as the Fourier transform X(jw) is the envelope function of the set of Fourier coefficients {a_k}, where $ prob(x) = \int_{x^-}^{x^+}f_X(x) dx $? -ag
- Sorry, I am not sure what you mean by "envelop function." -pm
Answer 2
Write it here.
Answer 3
Write it here.
