Consider two normal random variables $X$ and $Y$:
$$X\sim N(m_1,s_1), \qquad Y\sim N(m_2,s_2)$$
I want to find the real number $c$ for which we have:
$$P(X<c)=P(Y<c)$$
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I'll assume that by $s_1$ you mean the variance, so that $\sqrt{s_1}$ is the standard deviation. Since the cumulative distribution function of the normal distribution is one-to-one, what you need is only this: $$ \frac{c-m_1}{\sqrt{s_1}} = \frac{c-m_2}{\sqrt{s_2}}. $$ It follows that $$ \sqrt{s_2}(c-m_1) = \sqrt{s_1}(c-m_2) $$ $$ (\sqrt{s_2}-\sqrt{s_1})c = m_1\sqrt{s_2} - m_2\sqrt{s_1} $$ $$ c=\cdots $$ I'll let you do the rest. |
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Two obvious solutions are when $\lim_{x\rightarrow\infty}$ or when $\lim_{x\rightarrow-\infty}$. Then the solution exists independent of the parameters. One other special case is when the likelihood ratio $(f_1/f0)(x)$ is monotone increasing. In this case the solution exists if and only if when $\lim_{x\rightarrow\infty}$ or $\lim_{x\rightarrow-\infty}$, which is true iff when $s_1=s_2$. For an arbitrary solution refer to Michael Hardy's solution. |
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