# How to compute the value for $\mu$ for Normal Random Variable from the given probabilistic relation

Q. Let $$X \sim N(\mu,\sigma_1^2)$$ and $$Y \sim N(\mu,\sigma_2^2)$$ be independent random variables such that

$$P[$$2X+4Y$$\leq 10] + P[$$3X+Y$$\leq 9] = 1$$

$$P[$$2X-4Y$$\leq 6] + P[$$-3X+Y$$\geq 1] = 1$$

How do I compute the value for $$\mu$$?

My approach was to assume 4 Normal random variables $$A$$, $$B$$, $$C$$ and $$D$$ as ($$2X+4Y$$), ($$3X+Y$$), ($$2X-4Y$$) and ($$-3X+Y$$) respectively and then calculate Mean and variance for each of them in given variables.

Then converting the given relation for a standard normal random variable by subtracting with respective mean values and dividing with standard deviation both sides of the inequality so I can figure out values from the table.

Now I wanted to know if there any relation between let's say for example two random variable when the sum of their individual probabilities is $$1$$.

$$\mu_A = 2\mu_X + 4\mu_Y = 6\mu$$

$$\mu_B = 4\mu$$

$$\mu_C = -2\mu$$

$$\mu_D = -2\mu$$

Similarly calculating VARIANCE for each variable:


$$\sigma_A = \sqrt[]{(2^2\sigma_1^2 + 4^2\sigma_2^2)}$$ and so for others..

$$P[\frac{A-\mu_A}{\sigma_A} \leq \frac{10-\mu_A}{\sigma_A}] + P[\frac{B-\mu_B}{\sigma_B} \leq \frac{9-\mu_A}{\sigma_B}] = 1$$

$$P[ z \leq \frac{10-\mu_A}{\sigma_A}] + P[z \leq \frac{9-\mu_A}{\sigma_B}] = 1$$

How to proceed?

• Please replace 'pictures' by JaX. Jun 17, 2021 at 20:44
• @BruceET done sir :) Jun 18, 2021 at 12:11

For a normal random variable $$W$$ with mean $$\mu$$ and standard deviation $$\sigma$$, the probability of the event $$(W \leq \alpha)$$ can be expressed in terms of $$\Phi(\cdot)$$, the CDF of a standard normal random variable as $$P(W \leq \alpha) = \Phi\left( \frac{\alpha-\mu}{\sigma}\right).$$ Now, an important property of $$\Phi(\cdot)$$ is that $$\Phi(\alpha) + \Phi(\beta) = 1 \iff \alpha + \beta = 0.$$ I hope that you will have discovered as part of your "so for others" that $$\sigma_C = \sigma_A$$ and $$\sigma_D = \sigma_B$$. So, use these to get two different expressions for $$\mu$$ in terms of $$\sigma_A$$ and $$\sigma_B$$ and hence in terms of $$\sigma_1$$ and $$\sigma_2$$.