In an example, I am told that


is $n(0,T)$, i.e., which I took to mean normally distributed with mean of $0$ and variance of $T$.

Note: The value of $b$ and $T$ are unspecified in the problem.

I know the PDF and CDF of the normal and standard normal distributions and I didn't make the pattern recognition. Can you help me see how it is a normal distribution with mean of $0$ and variance of $T$?

Thanks in advance.

Update: Since I am in agreement with Chinny84 and Did that this function does not qualify as being normally distributed, I have chosen to interpret the author's notation as not meaning a normal distribution, but as a simplifying substitution rather than implying a normal distribution - despite the unfortunate use of notation that normally (pun a natural consequence) denotes the normal distribution. And that approach does work (I can't elaborate). The source is a tiny part of a larger problem statement on an exam. I'll award Chinny84 the answer for the help his answer will provide others who come after. Thanks again for your interest/help.

  • $\begingroup$ The RHS is a number so, how could it be "normally distributed"? What is your source? $\endgroup$ – Did Oct 5 '13 at 21:27

The above expression is not a normal which would have a form $$ \frac{1}{\sqrt{2\pi b}}\mathrm{e}^{-0.5\left(\frac{x-a}{b}\right)^2} $$

Where a is the mean and b is the standard dev.

The above is not even an exponential distribution since that would have the form $\frac{b}{1-\mathrm{e}^{-bT}}\mathrm{e}^{-bt}$ for [0,T).

To find mean and variance you would compute the corresponding moments ie integrals for a pdf. However, if you can see that it is clearly of Gaussian form then by inspection you can determine the mean and variance.

I always give warning when reading a response (like my one) and double check for yourself.

Hope it helps.


  • $\begingroup$ Also the coefficient for the exponential dist in my answer is to maintain the integral over that region is unity. $\endgroup$ – Chinny84 Oct 5 '13 at 21:43
  • $\begingroup$ Thanks Chinny84. I am leaving it open still to get additional responses, but yours is appreciated. $\endgroup$ – Joe Oct 5 '13 at 21:51
  • $\begingroup$ No problem. I am a firm believer in having different view points. Can you confirm that the expression you wrote was correct for future posters. $\endgroup$ – Chinny84 Oct 5 '13 at 22:04
  • $\begingroup$ No worries, and thank you for accepting my answer. All the best with everything. Rob $\endgroup$ – Chinny84 Oct 9 '13 at 12:28

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