Knowing that the random variable X obeys an exponential distribution with a parameter of 2, the expe

1 parameter , 1 2.

Random variable. Under different conditions, due to the influence of accidental factors, different values of various random variables may be taken, so they have uncertainty and randomness, but the probability that these values fall in a certain range is certain, and such variables are called random variables. Random variables can be either discrete or continuous.

For example, the measured value in the analytical test is a random variable with a probability value, and the value of the measured quantity may change randomly within a certain range, and the specific value cannot be determined before the measurement, but the result of the measurement is determined, and the measured value obtained by repeated measurement for many times has statistical regularity. The essential difference between the uncertainty of random variables and fuzzy variables is that the results of the latter are still uncertain, i.e., fuzzy.

1 parameter , 1 2.

The advantage of quantifying random events is that random phenomena can be studied by means of mathematical analysis. For example, the number of passengers waiting at a bus stop at a given time, the number of calls received by the exchange station in a certain time, the life of the light bulb, etc., are all examples of random variables.

When doing experiments, we are often interested in some function of the results relative to the results themselves. For example, when rolling dice, it is often the point and number of the two dice that are often cared for, and not really concerned with the actual result;

where >0 is a parameter of the distribution, often referred to as a rate parameter. That is, the number of times an event occurs per unit of time. Exponential distribution.

The interval is [0, ) if a random variable.

x is exponentially distributed, you can write: x e( ).

The expected exponential distribution is ex=1 and the variance dx=12

The probability density of x e( )x f(x) = e (-x), x>0, f(x) = 0, and x are others.

e(x)=∫0,∞)xf(x)dx=∫(0,∞)xλe^(-x)dx=1/λ，e(x²)=0,∞)x²f(x)dx=…=2/λ²

d(x)=e(x²)-e(x)]²1/λ²

The variance of the exponential distribution is 1

The random variable x obeys the exponential distribution of parameter 2, and it is expected that ex is equal to 1 2.

The expectation is equal to the integral of xf(x)dx on the x branch (where f(x) is the probability density of the random variable x), and for an exponential distribution that obeys the parameter a, the probability density is: f(x) = ae (-ax) when x is greater than or equal to 0, and f(x) = 0 when x is less than 0.

Then for the random variable x, which obeys the exponential distribution of any parameter a, ex=(the integral of x*ae (-ax) between 0 and positive infinity), i.e., ex=1 a, i.e., when the parameter in the problem is 2, the expected ex=1 2 of x.

p(y=0)=p(x>1)=e^(-1)

p(y=1)=p(x<=1)=1-e^(-1)dy=e^(-1)[1-e^(-1)]

An important feature of exponential functions is the absence of memory, which means that if a random variable is exponentially distributed, there is p(t>t+s|) when s,t>0t>t)=p(t>s)。That is, if t is the lifetime of a component and the component is known to use t hours, the total conditional probability that it will use at least s + t hours is equal to the probability that it will use at least s hours from the time of its inception.

Solution: Because Ling Tsai guesses that the random variable ruler x obeys the exponential distribution with the parameter of 1, Qi Wheel.

f(x)=e (-x)(at x>0).

and f(x)=0(x<=0).

e（x+e^(-2x))

e(x)+e(e(-2x))[let g(x)=e(-2x)]1+ f(x)g(x)dx(0 to infinity integral) 1+ e (-3x)dx

Mu Zhi Answer]: From the exponential distribution of the random variable x obeying the parameter of , the <> is then <>

And <> solution =.<>

Answer]: Exponential celery distribution

Knowing that x obeys the exponential distribution of the parameter of , and p=2p, the probability is calculated by the distribution function to obtain 1-e-1

2e-2/θ

Let u=e-1

Solve the one-element quadratic equation pin letter 2u2

u-1=0, you get u=1 2, or u=-1 round off the negative value, you get e-1 1 2 take the logarithm and get =1 ln2