There are two independent Gaussian R.Vs:

$U:N(-1,1)$ and $V:N(1,1)$

How do I go about finding the PDF of the following transformations?

  1. X = U+V

  2. T = (U+2V, U-2V)

  3. W = U (with 50% chance), V (with 50% chance)

Note: This is not a homework problem. Doing some self study and stuck.

For 1. I have some vague clue. Take CDF, Then Joint PDF...?

But not sure how to go about solving for 2. and 3. at all.

  • $\begingroup$ Finding the pdf of sums of RVs involves MGFs/characteristic functions if I am not mistaken. See [here][1]. [1]: en.wikipedia.org/wiki/… $\endgroup$ – BCLC Aug 25 '14 at 6:09
  • $\begingroup$ Guess for #3: $W = 1_E * U + 1_{E^{c}} * V$ where $1_E$ is the indicator function on event E where P(E) = 0.5. Then maybe MGFs again? $1_E, 1_{E^{c}}, U, V$ are all random variables. Maybe U and $1_E$ are independent so: en.wikipedia.org/wiki/Product_distribution After you get the pdfs of the products, get the pdfs of the sums $\endgroup$ – BCLC Aug 25 '14 at 6:10
  • $\begingroup$ en.wikipedia.org/wiki/… dartmouth.edu/~chance/teaching_aids/books_articles/… That's all I got. XD $\endgroup$ – BCLC Aug 25 '14 at 6:15
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    $\begingroup$ Thank you for all the resources. I had not thought about MGF approach at all. $\endgroup$ – Raaj Aug 27 '14 at 21:18

Since you have a vague idea for (1) I leave that one to you (hint: moment generating functions, and properties of sum of independent random variables regarding moment generating functions will be the way to go)

Problem 2

This problem becomes a lot easier if we prove the following proposition

If $X\sim N(\mu_{x},\sigma^{2})$ and $Y\sim N(\mu_{y},\sigma^{2})$ are independent, then $X+Y$ and $X-Y$ are independent

This has been proven quite a few times on forum (*). With this we see by properties of linear combos of independent random variables we have $$Var(U+2V)=Var(U)+4Var(V)$$ and $$Var(U-2V)=Var(U) +4Var(V)$$ Thus we can employ the above proposition which indicates that $U+2V$ and $U-2V$ are independent. If you recall, that the joint pdf of independent random variables, is just product of pdfs. To find pdfs of $U+2V$ and $U-2V$, your answer to (1) may help.

Problem 3

For this problem an important fact to know is that

if $X$ is an rv with pdf $f(x)$ and cdf $F(x)$, then $\frac{d}{dx}F(x)=f(x)$

And though this theorem in general may be hard to prove, when you have cdfs that are determined by integrating over pdf like is the case for normal rvs, this guaranteed by Fundamental Theorem of Calculus. Now to find distribution of W we are going to look at lens of its cdf and evoke the law of total probability. Well with this we have $$F_{W}(w)=P(W<w)=\frac{1}{2}P(U<w)+\frac{1}{2}P(V<w)$$ Thus to find pdf we will use above fact, $$f_{W}(w)=\frac{d}{dw}F_{W}(w)=\frac{d}{dw}\left(\frac{1}{2}F_{U}(w)+\frac{1}{2}F_{V}(w)\right)=\frac{1}{2}\frac{d}{dw}F_{U}(w)+\frac{1}{2}\frac{d}{dw}F_{V}(w)=\frac{1}{2}\left(f_{U}(w)+f_{V}(w)\right)$$


(*)Proof of proposition in problem (2): Prove that if $X$ and $Y$ are Normal and independent random variables, $X+Y$ and $X-Y$ are independent or Prove that if X and Y are Normal and independent random variables, X+Y and X−Y are independent−y-a

  • $\begingroup$ Hopefully this can help get you started $\endgroup$ – Kamster Aug 25 '14 at 6:59
  • $\begingroup$ Note for problem 2 I forgot that $2V$ variances changes thus $U$ and $2V$ don't have same variance so you could apply that theorem, what Did wrote should help you though $\endgroup$ – Kamster Aug 25 '14 at 17:22

In every exercise about normal random variables, it proves useful to reduce the problem to the study of a small number of independent standard normal random variables.

Here $U=-1+Y$ and $V=1+Z$ where $(Y,Z)$ is i.i.d. standard normal and one knows that, for every $(x,y,z)$, $x+yY+zZ$ is normal $N(x,y^2+z^2)$.

This should yield that $X$ is $N(0,2)$ and $T$ normal with covariance matrix $C=\begin{pmatrix}a^2&c\\ c&b^2\end{pmatrix}$ with $a^2=\mathrm{var}(U+2V)=5$, $b^2=\mathrm{var}(U-2V)=5$ and $c=\mathrm{cov}(U+2V,U-2V)=a^2-4b^2=-15$.

Question 3. is of a different nature, probably best solved noting that the PDFs are related through the identity $$f_W=\tfrac12f_U+\tfrac12f_V,$$ thus, for every $x$, $$f_W(x)=\frac12f_Y(x+1)+\frac12f_Z(x-1)=\frac12\frac1{\sqrt{2\pi}}\mathrm e^{-(x+1)^2/2}+\frac12\frac1{\sqrt{2\pi}}\mathrm e^{-(x-1)^2/2},$$ that is, $$f_W(x)=\frac12(\mathrm e^x+\mathrm e^{-x})\frac1{\sqrt{2\pi}}\mathrm e^{-1/2}\mathrm e^{-x^2/2}.$$ Note that $W$ is not normal.

  • $\begingroup$ just out of curiosity why do we have x+1, x-1 (for question 3) and not just x $\endgroup$ – Kamster Aug 25 '14 at 17:24
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    $\begingroup$ Because the densities of $C$ and $C+c$ are related by the identity $f_{C+c}(x)=f_C(x-c)$. $\endgroup$ – Did Aug 25 '14 at 17:48


(1) Pdf of sum of two independent random variables is the convolution of individual pdfs.

(2) Linear combination of independent Gaussian random variables is again a Gaussian variable.

(3) $E(c \cdot X)=c \cdot E(X)$

(4) $Var(c \cdot X)=c^2 \cdot Var(X)$


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