$E[X] = 1$, $E[(X – E[X])Z] = –2$, $E[(X – E[X])^2] = 4$,
where $E[X]$ denotes the expectation of random variable $X$. The values of $a, b$ are:
Key Concepts and Given Information
We are given a linear transformation $X = aZ + b$, where $Z$ is a standard normal random variable ($Z \sim N(0, 1)$). This means $E[Z] = 0$ and $Var(Z) = E[Z^2] = 1$. We need to find the constants $a$ and $b$ using the provided expectation conditions.
- $E[X] = 1$
- $E[(X – E[X])Z] = –2$
- $E[(X – E[X])^2] = 4$, which is equivalent to $Var(X) = 4$.
Solving for Constant b
From the condition $E[X] = 1$:
$ E[aZ + b] = 1 $
Using the linearity property of expectation:
$ aE[Z] + E[b] = 1 $
Since $E[Z] = 0$ and $E[b] = b$:
$ a(0) + b = 1 $
$ \implies b = 1 $
Solving for Constant a
From the condition $Var(X) = 4$:
$ Var(aZ + b) = 4 $
Using the property $Var(aZ + b) = a^2 Var(Z)$:
$ a^2 Var(Z) = 4 $
Since $Var(Z) = 1$:
$ a^2 (1) = 4 $
$ \implies a^2 = 4 $
This implies $a = 2$ or $a = -2$. We use the remaining condition to find the specific value of $a$.
Using the Second Expectation Condition
From the condition $E[(X – E[X])Z] = –2$:
Substitute $X = aZ + b$ and $E[X] = 1$. Since we found $b = 1$, $X = aZ + 1$ and $E[X] = 1$.
$ E[((aZ + 1) – 1)Z] = –2 $
Simplify the expression inside the expectation:
$ E[(aZ)Z] = –2 $
$ E[aZ^2] = –2 $
Using the linearity of expectation:
$ aE[Z^2] = –2 $
Since $E[Z^2] = Var(Z) = 1$:
$ a(1) = –2 $
$ \implies a = -2 $
Conclusion
We found $b = 1$ and $a = -2$. Therefore, the values are $a = -2$ and $b = 1$.
Important Questions from Random Variables
Consider the following probability density function for a random variable x.
(a) \(f_{1}(x)=1\ ;\ -1\le x\le 1\)
(b) \(f_{2}(x)=x\ ;\ 0\le x\le 1\)
(c) \(f_{3}(x)=(1-x)\ ;\ -1\le x\le 1\)
Arrange the above functions in terms of the increasing value of mean of random variable x.
Considering all the symbols with their usual meanings, match the following :
List - I List - II (a) \(f_{X}(x)\) (i) \(\displaystyle\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dx\) (b) \(\displaystyle\int_{-\infty}^{\infty}f_{X}(x)\,dx\) (ii) \(\displaystyle\int_{-\infty}^{a}f_{X}(x)\,dx\) (c) \(F_{X}(a)\) (iii) \(\displaystyle\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dy\) (d) \(f_{Y}(y)\) (iv) 1 Codes :
The density function of a random variable is given by \(p(x)=Ke^{-\frac{x^{2}}{2}}\) for \(-\infty \lt x \lt \infty\). The value of K should be :
If the odds in favour of any random event A are 5 ∶ 6, then the odds against the event are:
If random variable X follows binomial distribution with parameter n and p with mean 15 and variance 10, then the value of mode is
