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Let's say I have a normally distributed random variable $X$ with mean $t$ and standard deviation $t^2$. That is,

$X \sim N (t,t^2)$

Here, $t$ is a parameter.

If I set $t$ equal to 0, then the probability density at $X=0$, for instance becomes: $ \frac {1} {(0)( \sqrt{2 \pi})} e^{(\frac{0} {0})^2}$, that is, as $X$ tends to 0, the probability density tends to infinity.

Is this a valid probability density function, with $t=0$?

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    $\begingroup$ Yes, it's valid. Have a look at the delta function (e.g. en.wikipedia.org/wiki/Dirac_delta_function) and the Heaviside step function (e.g. en.wikipedia.org/wiki/Heaviside_step_function). $\endgroup$
    – LBogaardt
    Apr 29 '20 at 12:05
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    $\begingroup$ There is no 'density' when $t=0$ in the sense that the distribution is no longer absolutely continuous. $\endgroup$ Apr 29 '20 at 12:22
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    $\begingroup$ You have discovered that probability densities are not good mathematical objects to work with when considering limits or extreme cases of families of probability distributions. If you were to work with, say, the CDF or (even better) the characteristic function, you would encounter no difficulty with the case $t=0.$ $\endgroup$
    – whuber
    Apr 29 '20 at 12:23
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    $\begingroup$ Furthermore, $X=0$ is a zero probability event. $\endgroup$
    – Xi'an
    Apr 29 '20 at 13:19
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If $t=0$, you have normal distribution with mean 0 and variance 0. The latter means that your "random variable" $X$ is constant, the former means that this constant equals 0.

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  • $\begingroup$ Thank you! So if I have a prior distribution for $t$, and I want to find the probability density of $X=0$, can I simple integrate $p(X=0|t=0)p(t=0)$ from - infinity to + infinity to get the value, even though the value at t=0 is infinity? $\endgroup$ Apr 29 '20 at 11:40

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