You're not changing the parameters of the distribution, at least not in the typical sense.

Let's consider an exponential distribution, $X\sim exp(\beta)$ with PDF $f_X(x\vert \beta) = \dfrac{e^{\frac{-x}{\beta}}}{\beta}$.

If we have a shift in mean of 1 to get $Y\sim exp(\beta+1)$, then, we change the variance, as you note. However, let's change the PDF itself to $ f_Y(y\vert \beta) = \dfrac{e^{\frac{-(y-1)}{\beta}}}{\beta}$, then we get the same sort of shape but shifted. This is no longer an exponential distribution, but it has the same shape.

That is a location-only shift.

You can play this same sort of game with other PDFs by remembering that, for functions $f(x)$ in general, $f(x-a)$ is $f(x)$ shifted to the right by $a$ (so a left shift if $a<0$).

EDIT: I say that you're not changing parameters in the traditional sense because the exponential distribution traditionally only has the one parameter, but let's work with the $TPM(\beta,a)$ distribution with PDF $ f_X(x\vert \beta,a) = \dfrac{e^{\frac{-(x-1)}{\beta}}}{\beta}$. Then you get a location shift (and only a location shift) by changing the $a$ parameter and not touching $\beta$.

You're not changing the parameters of the distribution, at least not in the typical sense.

Let's consider an exponential distribution, $X\sim exp(\beta)$ with PDF $f_X(x\vert \beta) = \dfrac{e^{\frac{-x}{\beta}}}{\beta}$.

If we have a shift in mean of 1 to get $Y\sim exp(\beta+1)$, then, we change the variance, as you note. However, let's change the PDF itself to $ f_Y(y\vert \beta) = \dfrac{e^{\frac{-(y-1)}{\beta}}}{\beta}$; then we get the same sort of shape but shifted. This is no longer an exponential distribution, but it has the same shape.

That is a location-only shift.

You can play this same sort of game with other PDFs by remembering that, for functions $f(x)$ in general, $f(x-a)$ is $f(x)$ shifted to the right by $a$ (so a left shift if $a<0$).

EDIT: I say that you're not changing parameters in the traditional sense because the exponential distribution traditionally only has the one parameter, but let's work with the $TPM(\beta,a)$ distribution with PDF $ f_X(x\vert \beta,a) = \dfrac{e^{\frac{-(x-a)}{\beta}}}{\beta}$. Then you get a location shift (and only a location shift) by changing the $a$ parameter and not touching $\beta$.

You're not changing the parameters of the distribution, at least not in the typical sense.

Let's consider an exponential distribution, $X\sim exp(\beta)$ with PDF $f_X(x\vert \beta) = \dfrac{e^{\frac{-x}{\beta}}}{\beta}$.

If we have a shift in mean of 1 to get $Y\sim exp(\beta+1)$, then, we change the variance, as you note. However, let's change the PDF itself to $ f_Y(y\vert \beta) = \dfrac{e^{\frac{-(y-1)}{\beta}}}{\beta}$, then we get the same sort of shape but shifted. This is no longer an exponential distribution, but it has the same shape.

That is a location-only shift.

You can play this same sort of game with other PDFs by remembering that, for functions $f(x)$ in general, $f(x-a)$ is $f(x)$ shifted to the right by $a$ (so a left shift if $a<0$).

You're not changing the parameters of the distribution, at least not in the typical sense.

Let's consider an exponential distribution, $X\sim exp(\beta)$ with PDF $f_X(x\vert \beta) = \dfrac{e^{\frac{-x}{\beta}}}{\beta}$.

If we have a shift in mean of 1 to get $Y\sim exp(\beta+1)$, then, we change the variance, as you note. However, let's change the PDF itself to $ f_Y(y\vert \beta) = \dfrac{e^{\frac{-(y-1)}{\beta}}}{\beta}$, then we get the same sort of shape but shifted. This is no longer an exponential distribution, but it has the same shape.

That is a location-only shift.

You can play this same sort of game with other PDFs by remembering that, for functions $f(x)$ in general, $f(x-a)$ is $f(x)$ shifted to the right by $a$ (so a left shift if $a<0$).

EDIT: I say that you're not changing parameters in the traditional sense because the exponential distribution traditionally only has the one parameter, but let's work with the $TPM(\beta,a)$ distribution with PDF $ f_X(x\vert \beta,a) = \dfrac{e^{\frac{-(x-1)}{\beta}}}{\beta}$. Then you get a location shift (and only a location shift) by changing the $a$ parameter and not touching $\beta$.