1
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ Please replace 'pictures' by JaX. $\endgroup$
    – BruceET
    Jun 17, 2021 at 20:44
  • 1
    $\begingroup$ @BruceET done sir :) $\endgroup$
    – Pinalen
    Jun 18, 2021 at 12:11

1 Answer 1

2
$\begingroup$

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 $\sigma_2$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .