<p>Well assuming the random variables are absolutely continuous, you can use densities:</p>
<p><span class="math-container">$$f(u|x,v)=\frac{f(u,v,x)}{f(x,v)}=\frac{f(u,v)f(x)}{f(x,v)}=\frac{f(u|v)f(v)f(x)}{f(x,v)}=f(u|v),$$</span></p>
<p>where the second and last equalities use independence between <span class="math-container">$x$</span> and <span class="math-container">$(u,v)$</span>.</p>

CC BY-SA 4.0
2022-02-15T14:57:07.760Z
You may proceed analogously using mass functions if they are discrete
Golden_Ratio
CC BY-SA 4.0
2022-02-15T14:55:31.080Z
2022-02-15T14:55:31.080Z
Golden_Ratio
<p>If the vector <span class="math-container">$(u,v)$</span> is independent of the vector <span class="math-container">$x$</span>, then I would like to show that
<span class="math-container">$$E(u|x,v)= E(u|v)$$</span></p>
<p>The only thing I can derive from the definitions is that if <span class="math-container">$(u,v)$</span> is independent of <span class="math-container">$x$</span>, then <span class="math-container">$E( (u,v) | x)= E((u,v))$</span>.</p>
<p>I can no longer attack this problem!</p>
<p>Help</p>

CC BY-SA 4.0
2022-02-15T07:17:10.387Z
What level do you want your answer at? Discrete random variables? Everything with densities? The Kolmogorov formulation of a conditional expectation as a Radon-Nikodym derivative?
Michael Greinecker
<probability>
A question about conditional expectation involving independence
PSE
# A question about conditional expectation involving independence

(Asked by: username_0 on 2022-02-15)

If the vector $(u,v)$ is independent of the vector $x$, then I would like to show that
$$E(u|x,v)= E(u|v)$$

The only thing I can derive from the definitions is that if $(u,v)$ is independent of $x$, then $E( (u,v) | x)= E((u,v))$.

I can no longer attack this problem!

Help

**Tags:** probability

> **Comments:**
> **username_1**: What level do you want your answer at? Discrete random variables? Everything with densities? The Kolmogorov formulation of a conditional expectation as a Radon-Nikodym derivative?  

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## 1 answers

### Answer by username_2 on 2022-02-15. Score: 6 | Accepted answer

Well assuming the random variables are absolutely continuous, you can use densities:

$$f(u|x,v)=\frac{f(u,v,x)}{f(x,v)}=\frac{f(u,v)f(x)}{f(x,v)}=\frac{f(u|v)f(v)f(x)}{f(x,v)}=f(u|v),$$

where the second and last equalities use independence between $x$ and $(u,v)$.

> **Comments:**
> **username_2**: You may proceed analogously using mass functions if they are discrete  

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