We can transform one continuous multivariate distribution to another based on two chosen transformation functions, their inverses and derivatives. These course notes go through the process in detail, but to cut to the chase:

  • Start with two random variables $X_1$ and $X_2$.
  • Assume the associated bivariate probability density function is $f(x_1, x_2)$.
  • Choose two transformation functions $y_1(x_1, x_2)$ and $y_2(x_1, x_2)$.
  • Let the derived random variables be $Y_1 = y_1(X_1, X_2)$ and $Y_2 = y_2(X_1, X_2)$.
  • Assume the associated bivariate probability density function is $g(y_1, y_2$).
  • Assume the inverses of the two transformation functions are $x_1(y_1, y_2)$ and $x_2(y_1, y_2)$.
  • The relationship between $f(x_1, x_2)$ and $g(y_1, y_2)$ is: $$ g(y_1, y_2) = f(x_1(y_1, y_2), x_2(y_1, y_2) \cdot |J| $$

Where J is the determinant of the Jacobian matrix of $[x_1, x_2]$ with respect to $[y_1, y_2]$:

$$ J = \frac{∂(x_1, x_2)}{∂(y_1, y_2)} $$

  • As a special case, if $f(x1, x2)$ corresponds to a uniform distribution, the relationship is:

$$ g(y_1, y_2) = |J| $$

Suppose that $X_1$ and $X_2$ are continuous i.i.d. random variables distributed Uniform(0,1).

Let $Y_1 = X_1 + X_2$ and $Y_2 = X_1 - X_2$

Note that $Y_2$ can be anything since it's only $Y_1$ we care about.

As far as I can tell, this setup satisfies the requirements of the process outlined above. The transformation functions have inverses, they determine a bijective mapping between the domain of $f$ and the domain of $g$.

So it should be possible to derive first the joint distribution $g$ of $Y_1$ and $Y_2$ and then the marginal distribution of $Y_1$, which we know is triangular.

But when I go through the process that's not what I get. In fact, my joint distribution (which is just $|J|$) ends up being a constant $1/2$, which doesn't sum to $1$ over its domain and therefore is not even a valid density function.

I know that normally we would derive the sum of two Uniform random variables using convolution, but I'd like to know if it's possible to use the above process (or if not, I'd like to know why).

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    $\begingroup$ I'd say: Yes it's possible, and it should be absolutely equivalent to a convolution integral, in your case. $\endgroup$ – Matteo Mar 12 at 18:44
  • $\begingroup$ That's good news! I would love to see a worked example or at least a hint on where I'm going wrong because my calculation of $g$ is obviously not correct. $\endgroup$ – chris838 Mar 13 at 12:42
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    $\begingroup$ As soon as I can I will write down an example. For now what I'm wondering is: are you sure that in your solution you didn't just miss to correct limits of integration? Because this is the main contribution to the integral in this case. $\endgroup$ – Matteo Mar 13 at 12:53

Consider two independent random variables $X_1$ and $X_2$ with uniform distribution over the interval $[0,1]$. Thus we have $$ f_{X_i}(x) = \begin{cases} 1 & \mbox{if} \ x \in [0,1]\\ 0 & \mbox{otherwise} \end{cases} $$ for $i=1,2$. Consider also, e.g., the transformation \begin{eqnarray} Y_1 &=& X_1+X_2;\\ Y_2 &=& X_2; \end{eqnarray} Then, as stated by your notes, the joint distribution of $Y_1$ and $Y_2$ is given by $$ f_{Y_1,Y_2}(u,v) = \begin{cases} f_{X_1,X_2}(u-v,v) =f_{X_1}(u-v)\cdot f_{X_2}(v) & \mbox{if} \ \ (u,v)\in [0,1]\times[0,1]\\ 0 & \mbox{otherwise}, \end{cases} $$ where we used independence of $X_1$ and $X_2$.

Because you're interested only in the marginal distribution of $Y_1$ we get $$ f_{Y_1}(u) = \int_{0}^1 f_{X_1}(u-v)f_{X_2}(v) dv, $$ that is exactly the convolution you were expecting.

EDIT As you requested, I little bit more details. If you use the transformations \begin{eqnarray} Y_1 &=& g_1(X_1,X_2)\\ Y_2 &=& g_2(X_1,X_2) \end{eqnarray} you first need to invert them and get \begin{eqnarray} X_1 &=& h_1(Y_1,Y_2)\\ X_2 &=& h_2(Y_1,Y_2). \end{eqnarray} Only now, I would recommend, can you

  1. Calculate, from $h_1(u,v)$ and $h_2(u,v)$, $$J = \left|\begin{array}{cc} \frac{\partial h_1}{\partial u} & \frac{\partial h_1}{\partial v} \\ \frac{\partial h_2}{\partial u} & \frac{\partial h_2}{\partial v}\end{array} \right|$$
  2. Determine the joint distribution of $(Y_2,Y_1)$, that, limited to the correct domain, will be given by $$f_{Y_1,Y_2}(u,v) = |J|\cdot f_{X_1,X_2}(h_1(u,v),h_2(u,v)).$$

In my example \begin{eqnarray} X_1 &=& Y_1-Y_2\\ X_2 &=& Y_2 \end{eqnarray} are the inverse transformations, so that \begin{eqnarray} h_1(u,v) &=& u-v\\ h_2(u,v) &=& v; \end{eqnarray}

  • $\begingroup$ Thanks! This makes sense and I think I see what I did wrong. I think I need a variable substitution in limits of my integral to get to the form you have here. $\endgroup$ – chris838 Mar 14 at 12:47
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    $\begingroup$ @chris838 you're welcome! And: yes, but as you see it would be a little "convoluted" (...) to have a more complicated $Y_2$, since you have to marginalize it "out". $\endgroup$ – Matteo Mar 14 at 12:49
  • $\begingroup$ Yeh the $Y_2$ makes this much simpler in fact. I'm still stuck on how you get $f_{X_1,X_2}(u−v, v)$. I end up with $f_{X_1,X_2}((1/2)(u + v))$ - is that on the right track? $\endgroup$ – chris838 Mar 14 at 13:38
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    $\begingroup$ @chris838 Ok, I added some more details at the bottom. Are my steps clearer now? $\endgroup$ – Matteo Mar 14 at 18:54
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    $\begingroup$ Thanks this is amazing! $\endgroup$ – chris838 Mar 14 at 19:06

Yes, you can of course use that approach to get the joint density function, but I don't think your way would be easier to calculate in general.

Regarding your question on $g(y1, y2)$ not being a valid density function. You actually made a mistake here. $g(y_1, y_2)/f(x_1, x_2) = 1/|J|$ rather than $J$.

Just take an example of a single-variable density function ($C$ is the cumulative distribution function) $$ p(x) = \frac{dC}{dx}$ $$ $$ p(y) = \frac{dC}{dy} = \frac{dC}{dx} \cdot \frac{dx}{dy} = \frac{p(x)}{y'(x)} $$

  • $\begingroup$ Hmm, but the notes I refer to explicitly state $g(y_1,y_2) = |J|$ ?? $\endgroup$ – chris838 Mar 13 at 12:38
  • $\begingroup$ Possibly you're defining $J$ as $∂(y_1,y_2)/∂(x_1,x_2)$ whereas here I'm defining it as $∂(x_1,x_2)/∂(y_1,y_2)$ $\endgroup$ – chris838 Mar 13 at 12:48
  • $\begingroup$ Yes, you are right about the difference in our definition. However, in that case your calculation is wrong $\partial(y_1,y_2)/\partial(x_1, x_2)=2$ and $\partial(x_1, x_2)/\partial(y_1, y_2) = 1/2$. $\endgroup$ – MoonKnight Mar 13 at 16:44
  • $\begingroup$ Apologies - I was playing around with different transform functions. You're right and I've updated the question. $\endgroup$ – chris838 Mar 14 at 12:40
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    $\begingroup$ your new integral area in $(Y_1, Y_2)$ space is a diamond with 4 cornors (0,0), (1,1), (-1,1), (2, 0), which area is $2$. so it does sum up to $1$. $\endgroup$ – MoonKnight Mar 14 at 16:47

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