Math Help - Is a shift on a random variable distribution preserving?

1. Is a shift on a random variable distribution preserving?

Suppose you have a continuous random variable X with pdf $${{f}_{X}}(x)$$ and cdf $${{F}_{X}}(x)$$. Now look at the transformation Y = X + k, with k being a real number. It's easy to see that the distribution of Y is given by $${{F}_{Y}}(y)={{F}_{X}}(y-k)$$ and $${{f}_{Y}}(y)={{f}_{X}}(y-k)$$

Does this imply that Y has the same distribution as X? In other words, does a shift either to the left or to the right of the random variable preserve its distribution? It would seem obvious that the answer is "yes" (basically I'm taking the shape and moving it without distortions), but I've yet to see a reference on this yet. Any suggestions?

On the other hand, if the answer is "not necessarily", the only culprit can be a Pareto distribution, which has a funkier definition. However, I think it works even in this case (e.g. a shifted Pareto remains Pareto).

2. Hello,

It all depends on what you call "preserve a distribution". You can't keep the same name as each distribution has its own domain where values are taken. If you shift it, you won't have the same domain anymore.
But the shape of the distribution certainly remains the same as you just translate it. But that's all you can say.

3. Originally Posted by Moo
It all depends on what you call "preserve a distribution". You can't keep the same name as each distribution has its own domain where values are taken. If you shift it, you won't have the same domain anymore.
That's exactly what I'm looking for - keeping the same name.

Yes, the domain changes from [a,b] to [a+k, b+k]. However, I'm not sure that this is enough to destroy the idea that the name can't be kept. For example, if the domain is the real axis, a shift most certainly preserves the name (e.g. a shifted Normal remains Normal, same for Laplace, etc). Even on distributions defined on bounded intervals, there are cases where we get to keep the name (e.g. a Uniform on [0,1] shifted still remains Uniform, just that it's defined on the new interval; same for Beta). Can you find a conterexample?

4. What you are referring to is known canonically as a "location family" with the underlying operation a "location shift." Obviously you cannot say that a (for example) location-shifted Exponential(1) is still an Exponential(1), because the support is not the same.

This idea is similar to the idea of a "scale family" in which one looks at the effect of scalar multiplication by a positive quantity. The exponential and gamma families are scale families, for example, while the Normal and Cauchy families are location-scale families. In the case of the Normal, $\mu$ is the location parameter while $\sigma$ is the scale parameter.

But no, you don't get to keep the same name unless you are already dealing with a location family. I guess the uniform works too.

5. Originally Posted by theodds
What you are referring to is known canonically as a "location family" with the underlying operation a "location shift." Obviously you cannot say that a (for example) location-shifted Exponential(1) is still an Exponential(1), because the support is not the same.
Aha, thank you. So then, the question becomes "is it true that one can create a location family for every distribution?"

Or equivalently, is it true that any distribution D is invariant under a location shift transformation? If not, is there a counterexample available?

To me, there is a subtle connection here with group theory. Just like in group theory you have groups like the Reals that are closed under certain operations (like addition and multiplication), there should be something similar on the field of distributions, and the operation is the translation shift.

6. If you are dealing with a univariate random variable, I don't see why you can't add a constant to it. As you noted, if $X \sim F$ and $Y$ is equal in distribution to $X + k$ then $F_Y (y) = F(y - k)$. The resulting CDF is still valid, just shifted. You never get back the same distribution you started with though. You just get back something in the same (suitably defined) family of distributions, and it seems a bit of a tautology to say something like "the location family of a distribution is closed under location shifts." Make sure you are keeping distinct the distribution itself and its associated family (of which there may be more than one, e.g. a uniform(0, 1) is also a beta(1, 1)), which is a set of distributions.

What you might be getting at is something along the lines of the retention of the "shape" of the distribution. If you imagine graphing the CDF or PDF then this transformation always results in movement along the x-axis, so in that sense you get back something that has the same shape.

7. Originally Posted by theodds
You just get back something in the same (suitably defined) family of distributions
OK, this is what I would like to make sure it's true. Please check my math and let me know if I'm wrong.

Let $X$ be a Pareto r.v., with pdf ${{f}_{X}}(x)=a\theta^ax^{-a-1}$ defined for all $x\geq\theta$. Notice that the pdf of a Pareto is influenced by its domain. Now let's do the shift $Y=X+k$.

The shifted Pareto should have the pdf ${{f}_{Y}}(y)={{f}_{X}}(y-k)=a\theta^a(y-k)^{-a-1}$ defined for all $y\geq\theta+k$. Is $Y$ still Pareto? If it is, shouldn't its pdf be ${{f}_{Y}}(y)=a(\theta+k)^a(y-k)^{-a-1}$?

If this is correct, what I'm quoting from you above should be incorrect, and here's the counterexample. If the math is wrong, what's the silly mistake? It's driving me nuts

What you might be getting at is something along the lines of the retention of the "shape" of the distribution. If you imagine graphing the CDF or PDF then this transformation always results in movement along the x-axis, so in that sense you get back something that has the same shape.
Yep, that's what I mentioned in my original post. However, if the math is OK, then not even this might be true in general. The shifted Pareto should be distorted due to the $(\theta+k)^a$ rescaling, vs. the original $\theta^a$ factor.

8. Originally Posted by baudolino
OK, this is what I would like to make sure it's true. Please check my math and let me know if I'm wrong.

Let $X$ be a Pareto r.v., with pdf ${{f}_{X}}(x)=a\theta^ax^{-a-1}$ defined for all $x\geq\theta$. Notice that the pdf of a Pareto is influenced by its domain. Now let's do the shift $Y=X+k$.

The shifted Pareto should have the pdf ${{f}_{Y}}(y)={{f}_{X}}(y-k)=a\theta^a(y-k)^{-a-1}$ defined for all $y\geq\theta+k$. Is $Y$ still Pareto?

If it is, shouldn't its pdf be ${{f}_{Y}}(y)=a(\theta+k)^a(y-k)^{-a-1}$? Mr F says: Where has this come from? It's integral over $\, [\theta + k, \, + \infty)$ is not equal to 1. It's not a pdf.

If this is correct, what I'm quoting from you above should be incorrect, and here's the counterexample. If the math is wrong, what's the silly mistake? It's driving me nuts

Yep, that's what I mentioned in my original post. However, if the math is OK, then not even this might be true in general. The shifted Pareto should be distorted due to the $(\theta+k)^a$ rescaling, vs. the original $\theta^a$ factor.

9. Originally Posted by baudolino
OK, this is what I would like to make sure it's true. Please check my math and let me know if I'm wrong.

Let $X$ be a Pareto r.v., with pdf ${{f}_{X}}(x)=a\theta^ax^{-a-1}$ defined for all $x\geq\theta$. Notice that the pdf of a Pareto is influenced by its domain. Now let's do the shift $Y=X+k$.

The shifted Pareto should have the pdf ${{f}_{Y}}(y)={{f}_{X}}(y-k)=a\theta^a(y-k)^{-a-1}$ defined for all $y\geq\theta+k$. Is $Y$ still Pareto? If it is, shouldn't its pdf be ${{f}_{Y}}(y)=a(\theta+k)^a(y-k)^{-a-1}$?

If this is correct, what I'm quoting from you above should be incorrect, and here's the counterexample. If the math is wrong, what's the silly mistake? It's driving me nuts

Yep, that's what I mentioned in my original post. However, if the math is OK, then not even this might be true in general. The shifted Pareto should be distorted due to the $(\theta+k)^a$ rescaling, vs. the original $\theta^a$ factor.
This is what I meant by suitably defined. You don't get a Pareto back, you get a location-shifted Pareto back; a member of the location-scale Pareto family (the Pareto family is already a scale family, and the paremeter $k$ in this case is a location parameter).

10. I knew it was something silly. I don't need $(\theta+k)^a$ since the argument $y-k$ "translates" back $y$ onto $[\theta,\infty)$. Of course what I wrote is not a pdf. Even if I wanted to make it a pdf and introduce a scaling constant to force it to integrate to 1, the scaling factor comes out to be $(\theta/{\theta+k})^a$.

Multiply this by my - wrong - ${f_{Y}(y)}$ and...surprise, surprise

Thanks again. Sometimes I REALLY like to over-complicate things.

11. The support should change to $[\theta + k, \infty)$. The correct pdf should be

$\displaystyle f_Y (y) = \frac{a}{\theta} \left(\frac{y - k}{\theta}\right)^{-a - 1} \mbox{I}(y \in [\theta + k, \infty))$

which is what you had initially.

12. Originally Posted by theodds
The support should change to $[\theta + k, \infty)$. The correct pdf should be

$\displaystyle f_Y (y) = \frac{a}{\theta} \left(\frac{y - k}{\theta}\right)^{-a - 1} \mbox{I}(y \in [\theta + k, \infty))$

which is what you had initially.

Exactly. Yay, I just rediscovered Pareto distributions of the second kind. I need some sleep Thanks, all.