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How to determine $C$ if $f$ is a probability density function? $$f(x,t)=\dfrac {1}{C\sqrt{t}}e^{-\dfrac{{x}^2}{4t}}$$

Should I integrate the integral $$\int_{-\infty}^{\infty}f(x,t) \ \ dx$$

mesel
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pimbi
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2 Answers2

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By definition, a probability density function integral is $1$. That is:

$$\int_{-\infty}^{+\infty}f(x,t)\text{d}x = 1$$

In your case, the integral depends on the constant $C$.

So, you have to solve the following: $$\int_{-\infty}^{+\infty}f(x,t)\text{d}x = g(C) = 1$$

In your case, $g(C) = 2\frac{\sqrt{\pi}}{C}$, and hence $C = 2\sqrt {\pi}$

the_candyman
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  • $$ \int_{-\infty}^\infty f(x,t)\ dx=f(t) $$ – Tunk-Fey May 24 '14 at 09:10
  • @Tunk-Fey I wanted to emphasize the dependency on $C$ – the_candyman May 24 '14 at 09:16
  • But you cannot state $\displaystyle\int_{-\infty}^\infty f(x,t)\ dx=1$. – Tunk-Fey May 24 '14 at 09:20
  • why? If I fix the parameter $t$, then it must be $1$. Hence, it must be one for all $t$ – the_candyman May 24 '14 at 09:21
  • $f(x,t)$ is a bivariate PDF, how come "If I fix the parameter $t$, then it must be $1$"? – Tunk-Fey May 24 '14 at 09:28
  • I got your point. In this case $\int_{-\infty}^\infty f(x,t)\ dx=f(t,C)$ and $\int_{t_1}^{t_2} f(t,C)\ dt = (t_2-t_1)\frac{\sqrt{\pi}}{C} = 1$. But in this case, we need $t_1$ and $t_2$, and they can't be infinite, right? Anyway, if $t$ is simply a parameter and not a random variable, then my first approach is right. – the_candyman May 24 '14 at 09:34
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    Your answer seems correct but I chose to not answer this question since it's not completely stated. The OP didn't mention the support of $x$ and $t$. – Tunk-Fey May 24 '14 at 09:40
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    @Tunk-Fey "f(x,t) is a bivariate PDF" Actually,$f$ is not a bivariate PDF, rather, for each $t$, $f(\ ,t)$ is a univariate PDF, hence the approach in this solution is correct (and it yields $C=2\sqrt\pi$, not $\sqrt\pi$). – Did May 24 '14 at 09:40
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    @Did, you are right, I missed a $2$ – the_candyman May 24 '14 at 09:44
  • @Did I didn't say the approach is incorrect. I'm just questioning the statement $\displaystyle\int_{-\infty}^{+\infty}f(x,t)\ \text{d}x = 1$ in the answer. Indeed, $C=2\sqrt{\pi}$ if the support $-\infty<x<\infty$. – Tunk-Fey May 24 '14 at 09:45
  • @Tunk-Fey Sure you did (question it), and with no correct cause. This identity is the way to go (rather than the one in your first comment, which I fail to understand). – Did May 24 '14 at 09:47
  • @Did I would agree if the_candyman wrote: "In this case $\displaystyle\int_{-\infty}^{+\infty}f(x,t)\ \text{d}x = 1$ since ..." in the answer rather than what he writes right now. – Tunk-Fey May 24 '14 at 09:53
  • @Tunk-Fey I disagree. This answer is correct as it is and your three first comments are misleading, being based on the belief that $f$ is a bivariate PDF, which it is not. +1 for the "$2\sqrt\pi$ version" of the answer. – Did May 24 '14 at 10:00
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This answer is not really structural but can be enough if you have allready some knowledge about PDF-s.

Do you know that $$\frac{1}{\sigma\sqrt{2\pi}}e^{-\dfrac{x^{2}}{2\sigma^{2}}}$$ is a PDF for $\sigma>0$ linked with distribution $Norm(0,\sigma^{2})$?

Compare it with your $f(x,t)=\frac{1}{C\sqrt{t}}e^{-\dfrac{x^{2}}{4t}}$.

By setting $t=\frac{1}{2}\sigma^{2}$ we come to $\frac{2}{C\sigma\sqrt{2}}e^{-\dfrac{x^{2}}{2\sigma^{2}}}$ which can only be a PDF if $\frac{2}{C\sigma\sqrt{2}}=\frac{1}{\sigma\sqrt{2\pi}}$ or equivalently $C=2\sqrt{\pi}$.

drhab
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