Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can any one help me to calculate this function : $$f(y)=\max\limits_{\mu>0}[\exp(\frac{-n\mu^{2}}{\sigma^{2}})\exp(\frac{2\mu}{\sigma^{2}}\sum_{k=1}^{n}y_{k})]$$ where $y_{k}$ is random variable with normal distribution. $$$$ Thank you in advance.

Sorry, I had forgotten to put the second power of $\mu$ in first exponential(I modified it).

share|improve this question
    
0.0333325355322 –  Артём Царионов Nov 12 '12 at 19:18
add comment

2 Answers

To maximize $\exp(g(\mu)/\sigma^2)$ with $g(\mu)=-n\mu^2+2\mu s$ and $s=\sum\limits_{k=1}^ny_k$, one should maximise $g(\mu)$. Since $g'(\mu)=-2n\mu+2s$ is positive for $\mu\lt s/n$ and negative for $\mu\gt s/n$, $g(\mu)$ is maximal at $\mu=s/n$ and $f(y)=\exp(g(s/n)/\sigma^2)$. Since $g(s/n)=s^2/n$, $f(y)=\exp(s^2/(n\sigma^2))$.

share|improve this answer
add comment

If $S = \sum_{k=1}^n y_k$, you want to maximize $g(\mu) = \exp((2S - n)\mu/\sigma^2)$ over $\mu > 0$. Presumably $\sigma^2 > 0$. If $2S-n > 0$ the supremum is $+\infty$: $g(\mu) \to +\infty$ as $\mu \to +\infty$. If $2S-n = 0$, $g(\mu) = 1$ for all $\mu$. If $2S-n < 0$ the supremum is $1$, with $g(\mu) < 1$ for $\mu > 0$ and $g(\mu) \to 1$ as $\mu \to 1$.

Somehow I doubt that this is the answer to your real question, but I don't know what the real question is.

share|improve this answer
    
My main question is that whether the supremum depends on $\mu$ or not? –  sendra Nov 12 '12 at 18:54
2  
How can the $\sup$ over $\mu$ depend on $\mu$? Taking the $\sup$ over $\mu$ must remove functional dependence on $\mu$. –  copper.hat Nov 12 '12 at 19:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.