# I want to find the real number c for which we have: $P(X<c)=P(Y<c)$

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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## 2 Answers

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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