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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
    Commented Apr 29, 2020 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. $\endgroup$
    – StubbornAtom
    Commented Apr 29, 2020 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 ♦
    Commented Apr 29, 2020 at 12:23
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    $\begingroup$ Furthermore, $X=0$ is a zero probability event. $\endgroup$
    – Xi'an
    Commented Apr 29, 2020 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? $\endgroup$
    – Ishan Kashyap Hazarika
    Commented Apr 29, 2020 at 11:40

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