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For random variables (rv) $X$ and $Y$ on a space $\Omega$:

Assume the rv $X\sim f_0$ distributed and $Y(t)=c$ is a constant rv, i.e. $Y\sim \delta(t-c)$ using the $\delta$-distribution as a short notation. Using the convolution I get $X+c \sim f_0(y-c)$. (I think $X$ and $Y$ are independent as $\{t|Y(t) \in B\} = \Omega$ ($B$ is a Borel set in $\mathbb{R}$).)

Question: is this correct? If yes, ist there a good intuitive explanation for the shift?

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    $\begingroup$ Please see stats.stackexchange.com/questions/73623. $\endgroup$
    – whuber ♦
    Commented Jan 22 at 20:25
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    $\begingroup$ @whuber if I understand correctly, my results are correct and the question can be viewed as a special case of your linked question? $\endgroup$
    – Christoph
    Commented Jan 22 at 20:34
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    $\begingroup$ Indeed, yes. You can take $f_0$ to be the kernel and $c=x_1$ to be a single data point. Another approach begins by observing the convolution is the distribution of the sum $Y+X.$ $\endgroup$
    – whuber ♦
    Commented Jan 22 at 20:35

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