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Let X and Y be independent and continuous random variables. How do you solve for the conditional density of X+Y given X

dyuta
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2 Answers2

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The identities $$ P[X+Y\in\mathrm dz\mid X=x]=P[Y\in\mathrm dz-x\mid X=x]=P[Y\in\mathrm dz-x] $$ imply that the conditional density of $X+Y$ given $X$ is $$ f_{X+Y\mid X=x}(z)=f_Y(z-x). $$ Edit: A rigorous approach to find the conditional density $(g_x)_x$ of $X+Y$ conditionally on $X$ is to ask that, for every bounded function $u$, $$ E[u(X+Y)\mid X]=v(X),\qquad v(x)=\int u(z)g_x(z)\mathrm dz. $$ To do that, recall that, by definition, $E[u(X+Y)\mid X]=v(X)$ if and only if, for every bounded function $w$, $E[u(X+Y)w(X)]=E[v(X)w(X)]$, that is, $$ \iint u(x+y)w(x)f_X(x)f_Y(y)\mathrm dx\mathrm dy=\int v(x)w(x)f_X(x)\mathrm dx. $$ By the change of variable $(x,z)=(x,x+y)$, the LHS is also $$ \iint u(z)w(x)f_X(x)f_Y(z-x)\mathrm dx\mathrm dz, $$ hence, by identification, $$ v(x)=\int u(z)f_Y(z-x)\mathrm dz, $$ and, by identification again, everything works fine with $$ g_x(z)=f_Y(z-x). $$ Edit2: A rigorous approach to find the joint density $f$ of $(X+Y,X)$ is to ask that, for every bounded function $u$, $$ E[u(X+Y,X)]=\iint u(z,x)f(z,x)\mathrm dz\mathrm dx. $$ By definition and by the change of variable $(x,z)=(x,x+y)$, the LHS is also $$ \iint u(x+y,x)f_X(x)f_Y(y)\mathrm dx\mathrm dy=\iint u(z,x)f_X(x)f_Y(z-x)\mathrm dx\mathrm dz, $$ hence, by identification, $$ f(z,x)=f_X(x)f_Y(z-x). $$

Did
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  • Did-Thx. Not sure I understand the math behind the identity, though the intuition is clear as it's similar to discrete case. My understanding of conditional probability in the case of continuous random variables is that P[Y lies in Borel set A|X=x] =Integral over A of the density f(x,y)/g(x); f is the joint density of x,y and g is the marginal density of x. I also know that the density of Z=X+Y is the convolution of densities of X and Y respectively. Is there a way to use the type of definitions above, and possibly change of variables etc to get the result without using dz and taking limits. – dyuta Aug 22 '13 at 20:39
  • Rigorous answer added. – Did Aug 22 '13 at 20:53
  • Great Thx. I had a question on a more 'direct' approach - Suppose I could have found the joint density h of Z & X, and then tried to prove that h(z,x)/g(x) = k(z-x) where g and k are the marginals of X and Y respectively. Would that work. I am stuck trying to find the marginal of Z & X. In any case thanks for your help. – dyuta Aug 22 '13 at 21:04
  • I meant stuck trying to find the joint distribution of Z & X – dyuta Aug 22 '13 at 21:14
  • See Edit 2. $ $ – Did Aug 22 '13 at 21:24
  • Thanks Did that answers the problem completely - For now we can divide f(z,x) by the marginal density of X and get the result in the so called direct way. – dyuta Aug 22 '13 at 21:28
  • @Did If I wanted to extend this to three random variables, i.e. $P(X+Y+Z\vert X=x,Z=z)$, would I need mutual independence of all relevant random variables? Particularly, would a condition like $Y$ independent of $X$ and $Y$ Independent of $Z$ suffice? (it seems to me like the answer is no, it wouldn't suffice?) – user106860 May 18 '17 at 19:31
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    @user106860 Nope, as usual one would need Y independent of (X,Z), which is strictly stronger than simply Y independent of X and Y independent of Z. – Did May 18 '17 at 19:38
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Hint: $P(X+Y) = P(X+Y|X)P(X)$ and use the independence assumption to write $P(X+Y)$ as a convolution.

baudolino
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  • To be more precise with my question- assume the following definition of conditional density: if f(u,v) is the joint density of random variables (U,V) and g(v) is marginal density of V, then f(u,v)/g(v) is conditional density of U given V=v. Conditional distribution of U given V=v is the appropriate integral of this density. In your hint, does P(X+Y) represent the density of (X+Y)? And then are you saying that P(X+Y|X) = P(X+Y)/P(X)? – dyuta Aug 22 '13 at 19:28
  • Correct, it's the very definition of conditional probability. – baudolino Aug 22 '13 at 19:43
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    Shouldn't the answer be h(z,x)/k(x) where h is the joint density of X+Y and X, and k is the marginal density of X. My problem is how to find the joint density h. I am really trying to show that h(z,x)/k(x) = l(z-x) where l is the marginal density of Y. – dyuta Aug 22 '13 at 19:52
  • As in the discrete case. – dyuta Aug 22 '13 at 19:59