E(X) = 1/2 e^(2x) = ∫ e^(2x) dt
Understand the Problem
The question involves finding the mathematical expected value of an expression that includes the exponential function e raised to the power of x. It looks like it's comparing two forms of an integral or an equation involving e.
Answer
The equation equates to solving $\frac{1}{2} e^{2t} = e^{t} - 1$.
Answer for screen readers
The solution for $t$ is given by the condition that must be satisfied by substituting and solving the equations derived.
Steps to Solve
- Understanding the Expected Value
We denote the expected value of $e^X$ as $E[e^X]$. The equation provided compares $E[e^X]$ to an integral form.
- Set Up the Integral
We express the expected value using an integral: $$ E[e^X] = \int_0^\infty e^x f(x) , dx $$ where $f(x)$ is the probability density function.
- Equating the Expressions
From the problem, we have: $$ \frac{1}{2} e^{2x} = \int_0^t e^{x} , dt $$
- Integrate the Right Side
To solve for the integral: $$ \int e^{x} , dx = e^{x} + C $$
- Evaluate the Integral
Using the limits specified: $$ \int_0^t e^{x} , dx = e^{t} - 1 $$ So, equate it to $\frac{1}{2} e^{2t}$.
- Final Step: Solve for $t$
Now, we set the two equations equal: $$ \frac{1}{2} e^{2t} = e^{t} - 1 $$ Now solve this equation for $t$.
The solution for $t$ is given by the condition that must be satisfied by substituting and solving the equations derived.
More Information
This expected value is a fundamental concept in probability and statistics, relating to how much one can expect to obtain on average when performing a random variable experiment.
Tips
- Mistaking the integral limits or function definitions can lead to incorrect calculations. Always double-check the function being integrated.
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